Problems

2019 Paper 3 Q10

D: 1500.0 B: 1500.0

momentum oblique collision coefficient of restitution Newton's law of restitution deflection angle trigonometric identity optimisation smooth spheres

Two identical smooth spheres $P$ and $Q$ can move on a smooth horizontal table. Initially, $P$ moves with speed $u$ and $Q$ is at rest. Then $P$ collides with $Q$. The direction of travel of $P$ before the collision makes an acute angle $\alpha$ with the line joining the centres of $P$ and $Q$ at the moment of the collision. The coefficient of restitution between $P$ and $Q$ is $e$ where $e < 1$. As a result of the collision, $P$ has speed $v$ and $Q$ has speed $w$, and $P$ is deflected through an angle $\theta$.

Show that $$u \sin \alpha = v \sin(\alpha + \theta)$$ and find an expression for $w$ in terms of $v$, $\theta$ and $\alpha$.
Show further that $$\sin \theta = \cos(\theta + \alpha) \sin \alpha + e \sin(\theta + \alpha) \cos \alpha$$ and find an expression for $\tan \theta$ in terms of $\tan \alpha$ and $e$. Find, in terms of $e$, the maximum value of $\tan \theta$ as $\alpha$ varies.

Solution:

Since the impulse is along the line of centres, the velocities are as show in the diagram. Additionally, vertical velocity is unchanged, so: $v \sin (\theta + \alpha) = u \sin \alpha$ \begin{align*} \text{COM}(\rightarrow): && u \cos\alpha &= v \cos(\alpha + \theta) + w \\ \Rightarrow && w &= u \cos \alpha - v \cos (\alpha + \theta) \end{align*}
Since the approach speed (horizontally) is $u \cos \alpha$ the speed of separation is $e u \cos \alpha$, in particular $w - v \cos(\theta + \alpha) = e u \cos \alpha$ or $w = v \cos (\theta + \alpha) + e u \cos \alpha$. \begin{align*} && w &= w \\ && v \cos (\theta + \alpha) + e u \cos \alpha &= u \cos \alpha - v \cos (\alpha + \theta) \\ \Rightarrow && \frac{u \sin \alpha}{\sin (\alpha + \theta)} \cos (\theta + \alpha) + e u \cos \alpha &= u \cos \alpha - \frac{u \sin \alpha}{\sin (\alpha + \theta)} \cos (\alpha + \theta) \\ \Rightarrow && \sin \alpha \cos(\theta + \alpha) + e \sin (\alpha+\theta)\cos \alpha &= \sin(\alpha+\theta) \cos \alpha - \cos(\alpha+\theta)\sin \alpha \\ &&&= \sin ((\alpha+\theta)-\alpha) \\ &&&= \sin \theta \end{align*} as required. \begin{align*} && \sin \theta &= \cos(\theta+ \alpha)\sin \alpha + e \sin (\theta + \alpha) \cos \alpha \\ &&&= \cos \theta \cos \alpha \sin \alpha - \sin \theta \sin^2 \alpha + e \sin \theta \cos ^2 \alpha + e \cos \theta \sin \alpha \cos \alpha \\ \Rightarrow && \tan \theta \sec^2 \alpha &= \tan \alpha - \tan \theta \tan^2 \alpha + e \tan \theta + e \tan \alpha \\ \Rightarrow && \tan \theta (1 + \tan^2 \alpha+\tan^2 \alpha-e) &= \tan \alpha + e \tan \alpha \\ \Rightarrow && \tan \theta &= \frac{(1+e)\tan \alpha}{1-e + 2\tan^2 \alpha} \end{align*} We seek to maximise $y = \frac{x}{c+2x^2}$, \begin{align*} && \frac{\d y}{\d x} &= \frac{c+2x^2-4x^2}{(c+2x^2)^2} \\ &&&= \frac{c-2x^2}{(c+2x^2)^2} \end{align*} Therefore the maximum will occur at $x = \sqrt{c/2}$, ie $\tan \alpha = \sqrt{(1-e)/2}$ and theta will be $\displaystyle \frac{(1+e)\sqrt{(1-e)/2}}{2(1-e)} =\frac{1}{2\sqrt{2}} \frac{1+e}{\sqrt{1-e}}$

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2018 Paper 3 Q9

D: 1700.0 B: 1484.0

momentum collisions coefficient of restitution recurrence relation elastic wall two-body collision second order recurrence inequality

A particle $P$ of mass $m$ is projected with speed $u_0$ along a smooth horizontal floor directly towards a wall. It collides with a particle $Q$ of mass $km$ which is moving directly away from the wall with speed $v_0$. In the subsequent motion, $Q$ collides alternately with the wall and with $P$. The coefficient of restitution between $Q$ and $P$ is $e$, and the coefficient of restitution between $Q$ and the wall is 1. Let $u_n$ and $v_n$ be the velocities of $P$ and $Q$, respectively, towards the wall after the $n$th collision between $P$ and $Q$.

Show that, for $n\ge2$, \[ (1+k)u_{n} - (1-k)(1+e)u_{n-1} + e(1+k)u_{n-2} =0\,. \tag{$*$} \]
You are now given that $e=\frac12$ and $k = \frac1{34}$, and that the solution of $(*)$ is of the form \[ \phantom{(n\ge0)} u_n= A\left( \tfrac 7{10}\right)^n + B\left( \tfrac 5{7 }\right)^n \ \ \ \ \ \ (n\ge0) \,, \] where $A$ and $B$ are independent of $n$. Find expressions for $A$ and $B$ in terms of $u_0$ and $v_0$. Show that, if $0 < 6u_0 < v_0$, then $u_n$ will be negative for large $n$.

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2017 Paper 1 Q10

D: 1500.0 B: 1484.0

momentum collisions coefficient of restitution geometric series kinetic energy sequential collisions one-dimensional motion mass ratio

Particles $P_1$, $P_2$, $\ldots$ are at rest on the $x$-axis, and the $x$-coordinate of $P_n$ is $n$. The mass of $P_n$ is $\lambda^nm$. Particle $P$, of mass $m$, is projected from the origin at speed $u$ towards $P_1$. A series of collisions takes place, and the coefficient of restitution at each collision is $e$, where $0 < e <1$. The speed of $P_n$ immediately after its first collision is $u_n$ and the speed of $P_n$ immediately after its second collision is $v_n$. No external forces act on the particles.

Show that $u_1=\dfrac{1+e}{1+\lambda}\, u$ and find expressions for $u_n$ and $v_n$ in terms of $e$, $\lambda$, $u$ and $n$.
Show that, if $e > \lambda$, then each particle (except $P$) is involved in exactly two collisions.
Describe what happens if $e=\lambda$ and show that, in this case, the fraction of the initial kinetic energy lost approaches $e$ as the number of collisions increases.
Describe what happens if $\lambda e=1$. What fraction of the initial kinetic energy is \mbox{eventually} lost in this case?

Solution:

\begin{align*} \text{COM}: && mu &= mv + \lambda m u_1 \\ \Rightarrow && u &= v + \lambda u_1 \tag{1} \\ \text{NEL}: && e &= \frac{u_1-v}{u} \\ \Rightarrow && eu &= u_1 - v \tag{2} \\ (1)+(2) && (1+e)u &= (1+\lambda) u_1 \\ \Rightarrow && u_1 &= \frac{1+e}{1+\lambda}u \\ && v &= u_1 - eu \\ &&&= \frac{1+e - (1+\lambda)e}{1+\lambda} u \\ &&&= \frac{1-\lambda e}{1+\lambda}u \end{align*} Note that subsequent (first (and second)) are the same as these, therefore: \begin{align*} u_n &= \left ( \frac{1+e}{1+\lambda} \right)^n u \\ v_n &= \frac{1-\lambda e}{1+\lambda } u_n \\ &= \frac{1-\lambda e}{1+\lambda } \left ( \frac{1+e}{1+\lambda} \right)^n u \end{align*}
If $e > \lambda$ then $(1-\lambda e) > 1-e^2 > 0$ and \begin{align*} \frac{v_{n+1}}{v_n} &= \frac{1+e}{1+\lambda} > 1 \end{align*} So the particles are moving away from each other - hence no more collisions.
If $e = \lambda$ then $u_n = u$ and $v_n = (1-\lambda)u$ so all the particles end up moving at the same speed. \begin{align*} \text{initial k.e.} &= \frac12 m u^2 \\ \text{final k.e.} &= \frac12 m((1-e)u)^2 + \sum_{n = 1}^{\infty} \frac12 \lambda^n m ((1-e)u)^2 \\ &= \frac12mu^2(1-e)^2 \left ( \sum_{n=0}^{\infty} e^n \right) \tag{$e = \lambda$} \\ &= \frac12 mu^2(1-e)^2 \frac{1}{1-e} \\ &= \frac12m u^2 (1-e) \\ \text{change in k.e.} &= \frac12 m u^2 - \frac12m u^2 (1-e) \\ &= e\frac12m u^2 \end{align*} Ie the total energy lost approaches a fraction of $e$.
If $\lambda e = 1$, after the second collision the particle will be stationary. ie \begin{align*} \text{initial k.e.} &= \frac12 m u^2 \\ \text{k.e. after }n\text{ collisions} &= \frac12 \lambda^n m \left (\left ( \frac{1+e}{1+\lambda} \right)^n u \right)^2\\ &= \frac12 \lambda^n m \left ( \frac{1+\frac1{\lambda}}{1+\lambda} \right)^{2n} u&2\\ &= \frac12 \lambda^n m \left ( \frac{1+\frac1{\lambda}}{1+\lambda} \right)^{2n} u\\ &= \frac12 \lambda^n m \left ( \frac{1}{\lambda} \right)^{2n} u\\ &= \frac12 m \lambda^{-n} u\\ &\to 0 \end{align*} Eventually we lose all the kinetic energy.

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2017 Paper 3 Q11

D: 1700.0 B: 1484.0

momentum conservation of momentum kinetic energy sequential firing simultaneous firing inequality telescoping sum railway truck recoil projectile

A railway truck, initially at rest, can move forwards without friction on a long straight horizontal track. On the truck, $n$ guns are mounted parallel to the track and facing backwards, where $n>1$. Each of the guns is loaded with a single projectile of mass $m$. The mass of the truck and guns (but not including the projectiles) is $M$. When a gun is fired, the projectile leaves its muzzle horizontally with a speed $v-V$ relative to the ground, where $V$ is the speed of the truck immediately before the gun is fired.

All $n$ guns are fired simultaneously. Find the speed, $u$, with which the truck moves, and show that the kinetic energy, $K$, which is gained by the system (truck, guns and projectiles) is given by \[ K= \tfrac{1}{2}nmv^2\left(1 +\frac{nm}{M} \right) . \]
Instead, the guns are fired one at a time. Let $u_r$ be the speed of the truck when $r$ guns have been fired, so that $u_0= 0$. Show that, for $1\le r \le n\,$, \[ u_r - u_{r-1} = \frac{mv}{M+(n-r)m} \tag{$*$} \] and hence that $u_n < u\,$.
Let $K_r$ be the total kinetic energy of the system when $r$ guns have been fired (one at a time), so that $K_0 = 0$. Using $(*)$, show that, for $1\le r\le n\,$, \[ K_r -K_{r-1} = \tfrac 12 mv^2 + \tfrac12 mv (u_r-u_{r-1}) \] and hence show that \[ K_n = \tfrac{1}{2}nmv^2 +\tfrac{1}{2}mvu_n \,. \] Deduce that $K_n < K$.

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2016 Paper 1 Q10

D: 1484.0 B: 1500.0

momentum collisions coefficient of restitution Newton's law of restitution simultaneous collisions sequential collisions inequality conservation of momentum

Four particles $A$, $B$, $C$ and $D$ are initially at rest on a smooth horizontal table. They lie equally spaced a small distance apart, in the order $ABCD$, in a straight line. Their masses are $\lambda m$, $m$, $m$ and $m$, respectively, where $\lambda>1$. Particles $A$ and $D$ are simultaneously projected, both at speed $u$, so that they collide with $B$ and $C$ (respectively). In the following collision between $B$ and $C$, particle $B$ is brought to rest. The coefficient of restitution in each collision is $e$.

Show that $e = \dfrac {\lambda-1}{3\lambda+1}$ and deduce that $e < \frac 13\,$.
Given also that $C$ and $D$ move towards each other with the same speed, find the value of $\lambda$ and of $e$.

Solution:

Collision between A & B. Since the speed of approach is $u$ and the coefficient of restitution is $e$ we must have $v_B = v_A + eu$. \begin{align*} \text{COM}: && \lambda m u &= \lambda m (v_B - eu) + m v_B \\ \Rightarrow && v_B(\lambda + 1) &=\lambda (1+ e) u \\ \Rightarrow && v_B &= \frac{\lambda(1+ e)}{1+\lambda} u \end{align*}

Collision between A & B. Since the speed of approach is $u$ and the coefficient of restitution is $e$ we must have $v_D = v_C + eu$. \begin{align*} \text{COM}: && m(-u) &= mv_C + m(v_C + eu) \\ \Rightarrow && 2v_C &= -(1+e)u \\ \Rightarrow && v_C &= -\frac{1+e}{2} u \end{align*}

\begin{align*} \text{NEL}: && w_C &= e(v_B - v_C) \\ \text{COM}: && mv_B+ mv_C &= m w_C \\ \Rightarrow && w_C &= v_B + v_C\\ \Rightarrow && e(v_B - v_C) &= (v_B + v_C) \\ \Rightarrow && (1-e)v_B &= -(1+e)v_C \\ \Rightarrow && (1-e) \frac{\lambda(1+ e)}{1+\lambda} &= (1+e) \frac{1+e}{2} \\ \Rightarrow && 2\lambda - 2\lambda e &= 1+\lambda + e + \lambda e \\ \Rightarrow && (3\lambda +1)e &= \lambda - 1 \\ \Rightarrow && e &= \frac{\lambda -1}{3\lambda + 1} \\ &&&< \frac{\lambda - 1 + \frac{4}{3}}{3\lambda + 1} \\ &&& = \frac13 \end{align*}
Since they move towards each other at the same speed $w_C = - v_D$ \begin{align*} && w_C &= - v_D \\ \Rightarrow && v_B + v_C &= -(v_C+eu) \\ \Rightarrow && -eu &= v_B +2v_C \\ &&&= \frac{\lambda(1+ e)}{1+\lambda} u -(1+e)u \\ \Rightarrow && 1 &= \frac{\lambda(1+e)}{1+\lambda} \\ \Rightarrow && 1+\lambda &= \lambda \left ( 1 + \frac{\lambda -1}{3\lambda+1} \right) \\ &&&= \lambda \frac{4\lambda}{3\lambda +1} \\ \Rightarrow && 1+4\lambda + 3\lambda^2 &= 4\lambda^2 \\ \Rightarrow && 0 &= \lambda^2 - 4\lambda - 1 \\ \Rightarrow && \lambda &= \frac{4 \pm \sqrt{20}}{2} \\ &&&= 2\pm \sqrt{5} \\ \Rightarrow && \lambda &= 2 + \sqrt{5} \\ && e &= \frac{1+\sqrt{5}}{7+3\sqrt{5}} \\ &&&=\sqrt{5}-2 \end{align*}

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2016 Paper 2 Q9

D: 1600.0 B: 1473.6

momentum conservation of momentum energy conservation bullet and block resistance force relative motion work-energy theorem collisions

A small bullet of mass $m$ is fired into a block of wood of mass $M$ which is at rest. The speed of the bullet on entering the block is $u$. Its trajectory within the block is a horizontal straight line and the resistance to the bullet's motion is $R$, which is constant.

The block is fixed. The bullet travels a distance $a$ inside the block before coming to rest. Find an expression for $a$ in terms of $m$, $u$ and $R$.
Instead, the block is free to move on a smooth horizontal table. The bullet travels a distance $b$ inside the block before coming to rest relative to the block, at which time the block has moved a distance $c$ on the table. Find expressions for $b$ and $c$ in terms of $M$, $m$ and $a$.

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2014 Paper 1 Q10

D: 1500.0 B: 1500.0

momentum collisions coefficient of restitution energy conservation bouncing ball stacked balls free fall conservation of momentum

A uniform spherical ball of mass $M$ and radius $R$ is released from rest with its centre a distance $H+R$ above horizontal ground. The coefficient of restitution between the ball and the ground is $e$. Show that, after bouncing, the centre of the ball reaches a height $R+He^2$ above the ground.
A second uniform spherical ball, of mass $m$ and radius $r$, is now released from rest together with the first ball (whose centre is again a distance $H+R$ above the ground when it is released). The two balls are initially one on top of the other, with the second ball (of mass $m$) above the first. The two balls separate slightly during their fall, with their centres remaining in the same vertical line, so that they collide immediately after the first ball has bounced on the ground. The coefficient of restitution between the balls is also $e$. The centre of the second ball attains a height $h$ above the ground. Given that $R=0.2$, $r=0.05$, $H=1.8$, $h=4.5$ and $e=\frac23$, determine the value of $M/m$.

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2013 Paper 2 Q11

D: 1600.0 B: 1500.0

momentum collisions coefficient of restitution Newton's law of restitution sequential collisions three particles inequality conservation of momentum

Three identical particles lie, not touching one another, in a straight line on a smooth horizontal surface. One particle is projected with speed $u$ directly towards the other two which are at rest. The coefficient of restitution in all collisions is $e$, where $0 < e < 1\,$.

Show that, after the second collision, the speeds of the particles are $\frac12u(1-e)$, $\frac14u (1-e^2)$ and $\frac14u(1+e)^2$. Deduce that there will be a third collision whatever the value of $e$.
Show that there will be a fourth collision if and only if $e$ is less than a particular value which you should determine.

Solution:

First Collision:

By NEL, $v_2 = v_1 + eu$, so \begin{align*} \text{COM}: && mu &= mv_1 + m(v_1 + eu) \\ \Rightarrow && 2mv_1 &= mu(1-e) \\ \Rightarrow && v_1 &= \frac12 u(1-e) \\ && v_2 &= \frac12 u(1-e) + eu \\ &&&= \frac12 u(1+e) \end{align*} The second collision is identical to the first except replacing $u$ with $\frac12u(1+e)$, therefore after that collision: \begin{align*} && \text{first particle} &= \frac12 u(1-e) \\ && \text{second particle} &= \frac12 \left (\frac12 u(1+e) \right)(1-e) \\ &&&= \frac14 u(1-e^2) \\ && \text{third particle} &= \frac12 \left (\frac12 u(1+e) \right)(1+e) \\ &&&= \frac14 u(1+e)^2 \end{align*} After all these collisions, all particles are moving in the same direction (since they all have positive velocity), but the first particle is now travelling faster than the second particle (as $\frac12(1-e) < 1$). Therefore they will collide again.
The third collision:

The speed of approach will be $\frac12u(1-e) - \frac14u(1-e^2) = \frac14u(1-e)(2 - (1+e)) = \frac14 u(1-e)^2$, therefore by NEL, $w_2 = w_1 + \frac14ue(1-e)^2$ \begin{align*} \text{COM}: && m\frac12u(1-e) + m \frac14u(1-e^2) &= mw_1 + m\left (w_1 + \frac14ue(1-e)^2 \right) \\ \Rightarrow && \frac14u(1-e)(2+(1+e)) &= 2w_1 + \frac14ue(1-e)^2 \\ \Rightarrow && 2w_1 &= \frac14u(1-e)(3+e)-\frac14ue(1-e)^2 \\ &&&= \frac14u(1-e)(3+e-e(1-e)) \\ &&&= \frac14u(1-e)(3+e^2) \\ \Rightarrow && w_1 &= \frac18 u(1-e)(3+e^2) \\ && w_2 &= \frac18 u(1-e)(3+e^2) + \frac14ue(1-e)^2 \\ &&&= \frac18u(1-e)(3+e^2+2e(1-e)) \\ &&&= \frac18u(1-e)(3+2e-e^2) \\ &&&= \frac18u(1-e)(1+e)(3-e) \\ \end{align*} A fourth collision is possible, iff \begin{align*} && \frac18u(1-e)(1+e)(3-e)&> \frac14 u(1+e)^2 \\ \Leftrightarrow && (1-e)(3-e)&> 2 (1+e) \\ \Leftrightarrow &&3-4e-e^2&> 2+2e \\ \Leftrightarrow &&1-5e-e^2&>0 \\ \Leftrightarrow && e &< 3-\sqrt{2} \end{align*}

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2012 Paper 2 Q11

D: 1600.0 B: 1484.9

momentum collisions perfectly inelastic collision sequential collisions kinetic energy loss summation proof

A small block of mass $km$ is initially at rest on a smooth horizontal surface. Particles $P_1$, $P_2$, $P_3$, $\ldots$ are fired, in order, along the surface from a fixed point towards the block. The mass of the $i$th particle is $im$ ($i = 1, 2, \ldots$)and the speed at which it is fired is $u/i\,$. Each particle that collides with the block is embedded in it. Show that, if the $n$th particle collides with the block, the speed of the block after the collision is \[ \frac{2nu}{2k +n(n+1)}\,. \] In the case $2k = N(N+1)$, where $N$ is a positive integer, determine the number of collisions that occur. Show that the total kinetic energy lost in all the collisions is \[ \tfrac12 mu^2\bigg( \sum_{n=2}^{N+1} \frac 1 n \bigg)\,. \]

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2011 Paper 1 Q10

D: 1516.0 B: 1484.0

momentum elastic collision restitution free fall energy conservation vertical motion bouncing particle one-dimensional collision

A particle, $A$, is dropped from a point $P$ which is at a height $h$ above a horizontal plane. A~second particle, $B$, is dropped from $P$ and first collides with $A$ after $A$ has bounced on the plane and before $A$ reaches $P$ again. The bounce and the collision are both perfectly elastic. Explain why the speeds of $A$ and $B$ immediately before the first collision are the same. The masses of $A$ and $B$ are $M$ and $m$, respectively, where $M>3m$, and the speed of the particles immediately before the first collision is $u$. Show that both particles move upwards after their first collision and that the maximum height of $B$ above the plane after the first collision and before the second collision is \[ h+ \frac{4M(M-m)u^2}{(M+m)^2g}\,. \]

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2011 Paper 2 Q9

D: 1600.0 B: 1484.0

momentum collisions coefficient of restitution Newton's law of restitution conservation of momentum wall collision sequential collisions inequality

Two particles, $A$ of mass $2m$ and $B$ of mass $m$, are moving towards each other in a straight line on a smooth horizontal plane, with speeds $2u$ and $u$ respectively. They collide directly. Given that the coefficient of restitution between the particles is $e$, where $0 < e \le 1$, determine the speeds of the particles after the collision. After the collision, $B$ collides directly with a smooth vertical wall, rebounding and then colliding directly with $A$ for a second time. The coefficient of restitution between $B$ and the wall is $f$, where $0 < f \le 1$. Show that the velocity of $B$ after its second collision with $A$ is \[ \tfrac23 (1-e^2)u - \tfrac13(1-4e^2)fu \] towards the wall and that $B$ moves towards (not away from) the wall for all values of $e$ and $f$.

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2010 Paper 1 Q11

D: 1500.0 B: 1484.1

momentum coefficient of restitution circular groove repeated collisions direction reversal conservation of momentum geometric series Newton's law of restitution

Two particles of masses $m$ and $M$, with $M>m$, lie in a smooth circular groove on a horizontal plane. The coefficient of restitution between the particles is $e$. The particles are initially projected round the groove with the same speed $u$ but in opposite directions. Find the speeds of the particles after they collide for the first time and show that they will both change direction if $2em> M-m$. After a further $2n$ collisions, the speed of the particle of mass $m$ is $v$ and the speed of the particle of mass $M$ is $V$. Given that at each collision both particles change their directions of motion, explain why \[ mv-MV = u(M-m), \] and find $v$ and $V$ in terms of $m$, $M$, $e$, $u$ and $n$.

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2010 Paper 2 Q10

D: 1600.0 B: 1516.0

momentum elastic collision conservation of energy conservation of momentum inequality optimisation logarithms chain of collisions

In an experiment, a particle $A$ of mass $m$ is at rest on a smooth horizontal table. A particle $B$ of mass $bm$, where $b >1$, is projected along the table directly towards $A$ with speed $u$. The collision is perfectly elastic. Find an expression for the speed of $A$ after the collision in terms of $b$ and $u$, and show that, irrespective of the relative masses of the particles, $A$ cannot be made to move at twice the initial speed of $B$.
In a second experiment, a particle $B_1$ is projected along the table directly towards $A$ with speed $u$. This time, particles $B_2$, $B_3$, $\ldots\,$, $B_n$ are at rest in order on the line between $B_1$ and $A$. The mass of $B_i$ ($i=1$, $2$, $\ldots\,$, $n$) is $\lambda^{n+1-i}m$, where $\lambda>1$. All collisions are perfectly elastic. Show that, by choosing $n$ sufficiently large, there is no upper limit on the speed at which $A$ can be made to move. In the case $\lambda=4$, determine the least value of $n$ for which $A$ moves at more than $20u$. You may use the approximation $\log_{10}2 \approx 0.30103$.

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2010 Paper 3 Q11

D: 1700.0 B: 1469.5

momentum collisions friction Newton's second law relative motion bullet and block kinematics deceleration energy conservation embedding

A bullet of mass $m$ is fired horizontally with speed $u$ into a wooden block of mass $M$ at rest on a horizontal surface. The coefficient of friction between the block and the surface is $\mu$. While the bullet is moving through the block, it experiences a constant force of resistance to its motion of magnitude $R$, where $R>(M+m)\mu g$. The bullet moves horizontally in the block and does not emerge from the other side of the block.

Show that the magnitude, $a$, of the deceleration of the bullet relative to the block while the bullet is moving through the block is given by \[ a= \frac R m + \frac {R-(M+m)\mu g}{M}\, . \]
Show that the common speed, $v$, of the block and bullet when the bullet stops moving through the block satisfies \[ av = \frac{Ru-(M+m)\mu gu}M\,. \]
Obtain an expression, in terms of $u$, $v$ and $a$, for the distance moved by the block while the bullet is moving through the block.
Show that the total distance moved by the block is \[ \frac{muv}{2(M+m)\mu g}\,. \]

Describe briefly what happens if $R< (M+m)\mu g$.

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2009 Paper 1 Q11

D: 1500.0 B: 1500.0

momentum collisions coefficient of restitution kinetic energy loss one-dimensional collision conservation of momentum Newton's law of restitution algebraic manipulation

Two particles move on a smooth horizontal table and collide. The masses of the particles are $m$ and $M$. Their velocities before the collision are $u{\bf i}$ and $v{\bf i}\,$, respectively, where $\bf i$ is a unit vector and $u>v$. Their velocities after the collision are $p{\bf i}$ and $q{\bf i}\,$, respectively. The coefficient of restitution between the two particles is $e$, where $e<1$.

Show that the loss of kinetic energy due to the collision is \[ \tfrac12 m (u-p)(u-v)(1-e)\,, \] and deduce that $u\ge p$.
Given that each particle loses the same (non-zero) amount of kinetic energy in the collision, show that \[ u+v+p+q=0\,, \] and that, if $m\ne M$, \[ e= \frac{(M+3m)u + (3M+m)v}{(M-m)(u-v)}\,. \]

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