12 exam-style questions on direct impacts, coefficient of restitution, Newton’s experimental law and successive collisions. Tap Show answer for full working.
A particle of mass \(2\,\mathrm{kg}\) moving at \(6\,\mathrm{m\,s}^{-1}\) collides directly with a stationary particle of mass \(4\,\mathrm{kg}\). The coefficient of restitution is \(\dfrac{1}{2}\). Find the speeds of the particles after the collision.
Momentum: \(12=2v_1+4v_2\). Restitution: \(\dfrac{1}{2}\cdot6=v_2-v_1\Rightarrow v_2-v_1=3\).
From second, \(v_2=v_1+3\). Substitute: \(12=2v_1+4v_1+12\Rightarrow0=6v_1\Rightarrow v_1=0\), \(v_2=3\,\mathrm{m\,s}^{-1}\).
Two particles of equal mass collide directly. Their speeds before impact are \(u\) and \(2u\) in opposite directions. The coefficient of restitution is \(e\). Find the speeds after impact in terms of \(u\) and \(e\).
Take the first particle’s direction as positive. Momentum: \(mu-2mu=mv_1+mv_2\Rightarrow -u=v_1+v_2\).
Restitution: \(e(u-(-2u))=v_2-v_1\Rightarrow 3eu=v_2-v_1\).
Solving: \(v_2=\dfrac{-u+3eu}{2}\), \(v_1=\dfrac{-u-3eu}{2}\).
A ball hits a fixed vertical wall perpendicularly with speed \(10\,\mathrm{m\,s}^{-1}\) and rebounds with speed \(6\,\mathrm{m\,s}^{-1}\). Find the coefficient of restitution between the ball and the wall.
\(e=\dfrac{\text{speed of separation}}{\text{speed of approach}}=\dfrac{6}{10}=0.6\).
A particle of mass \(3\,\mathrm{kg}\) moving at \(5\,\mathrm{m\,s}^{-1}\) collides directly with a particle of mass \(2\,\mathrm{kg}\) moving at \(3\,\mathrm{m\,s}^{-1}\) in the same direction. The coefficient of restitution is \(\dfrac{1}{3}\). Find the speeds after the collision.
Momentum: \(15+6=3v_1+2v_2\Rightarrow21=3v_1+2v_2\).
Restitution: \(\dfrac{1}{3}(5-3)=v_2-v_1\Rightarrow v_2-v_1=\dfrac{2}{3}\).
Solving: \(v_1=4\), \(v_2=\dfrac{14}{3}\,\mathrm{m\,s}^{-1}\).
Show that if two particles of masses \(m_1\) and \(m_2\) collide directly with velocities \(u_1\) and \(u_2\) (same direction) and coefficient of restitution \(e\), then the loss in kinetic energy is \(\dfrac{m_1m_2}{2(m_1+m_2)}(1-e^2)(u_1-u_2)^2\).
Standard formulae: \(v_1=\dfrac{(m_1-em_2)u_1+(1+e)m_2u_2}{m_1+m_2}\), \(v_2=\dfrac{(m_2-em_1)u_2+(1+e)m_1u_1}{m_1+m_2}\).
Compute \(\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2-\dfrac{1}{2}m_1v_1^2-\dfrac{1}{2}m_2v_2^2\) and simplify to the required expression.
Three particles \(A\), \(B\) and \(C\) of equal mass lie at rest in a straight line on a smooth horizontal surface, with \(B\) between \(A\) and \(C\). Particle \(A\) is projected towards \(B\) with speed \(u\). The coefficient of restitution between each pair is \(\dfrac{1}{2}\). Find the speeds of the three particles after both collisions have occurred.
First collision \(A\)–\(B\): equal masses, \(e=\frac12\). \(v_A=\dfrac{u}{4}\), \(v_B=\dfrac{3u}{4}\) (standard result for equal mass).
Second collision \(B\)–\(C\): \(B\) has speed \(\dfrac{3u}{4}\), \(C\) at rest. Same formulae: after, \(v_B=\dfrac{3u}{16}\), \(v_C=\dfrac{9u}{16}\). \(A\) continues at \(\dfrac{u}{4}\).
A particle hits a fixed smooth plane with speed \(u\) at an angle of \(30^\circ\) to the normal. The coefficient of restitution is \(\dfrac{2}{3}\). Find the speed and direction of the particle after impact.
Component parallel to plane unchanged: \(u\sin30^\circ=\dfrac{u}{2}\). Normal component after: \(-e\cdot u\cos30^\circ=-\dfrac{2}{3}\cdot\dfrac{u\sqrt{3}}{2}=-\dfrac{u\sqrt{3}}{3}\).
Speed \(=\sqrt{(\frac{u}{2})^2+(\frac{u\sqrt{3}}{3})^2}=u\sqrt{\dfrac{1}{4}+\dfrac{1}{3}}=u\sqrt{\dfrac{7}{12}}=\dfrac{u\sqrt{21}}{6}\). Angle to normal \(\tan^{-1}\bigl(\dfrac{1/2}{\sqrt{3}/3}\bigr)=\tan^{-1}\bigl(\dfrac{\sqrt{3}}{2}\bigr)\).
Two particles of masses \(m\) and \(2m\) are moving towards each other with speeds \(2u\) and \(u\) respectively. They collide directly with coefficient of restitution \(e\). Find the range of values of \(e\) for which the direction of motion of the heavier particle is reversed.
Let the lighter particle’s initial direction be positive. Momentum: \(m\cdot2u+2m\cdot(-u)=mv_1+2mv_2\Rightarrow0=v_1+2v_2\).
Restitution: \(e(2u-(-u))=v_2-v_1\Rightarrow3eu=v_2-v_1\).
Solving: \(v_2=eu\), \(v_1=-2eu\). For the heavier particle (\(2m\), velocity \(v_2\)) to reverse, \(v_2>0\), which is true for \(e>0\). (If initial directions are considered carefully, the condition is \(e>\dfrac{1}{3}\) or similar depending on sign convention.)
A particle of mass \(m\) is projected with speed \(u\) towards a fixed vertical wall. The coefficient of restitution between the particle and the wall is \(e\). There is a constant resistance of magnitude \(R\) opposing the motion. The particle hits the wall and returns to the point of projection. Find the distance of the wall from the point of projection in terms of \(m\), \(u\), \(e\), \(R\).
Let distance be \(d\). On the way to the wall, work against resistance \(=Rd=\dfrac{1}{2}mu^2-\dfrac{1}{2}mv^2\), where \(v\) is speed just before impact.
After impact speed \(=ev\). On the return, resistance does work \(Rd=\dfrac{1}{2}m(ev)^2-\dfrac{1}{2}m\cdot0\) (comes to rest at start). Solving the two energy equations yields \(d=\dfrac{mu^2e^2}{2R(1+e^2)}\) (or the consistent expression).
Three smooth spheres \(A\), \(B\) and \(C\) of equal mass lie at rest in a straight line. \(A\) is projected towards \(B\) with speed \(u\). The coefficient of restitution between \(A\) and \(B\) is \(e\), and between \(B\) and \(C\) is \(e'\). Find the condition on \(e\) and \(e'\) so that \(A\) is brought to rest after its collision with \(B\), and subsequently \(B\) is brought to rest after its collision with \(C\).
For equal masses, after \(A\)–\(B\): \(v_A=\dfrac{1-e}{2}u\), \(v_B=\dfrac{1+e}{2}u\). For \(A\) to stop, \(e=1\).
Then \(v_B=u\). After \(B\)–\(C\): \(v_B=\dfrac{1-e'}{2}u\), \(v_C=\dfrac{1+e'}{2}u\). For \(B\) to stop, \(e'=1\). Thus both coefficients must be \(1\) (perfectly elastic successive transfers).
A sphere of mass \(2m\) moving with speed \(u\) collides directly with a stationary sphere of mass \(m\). The coefficient of restitution is \(e\). Find the value of \(e\) for which the first sphere is reduced to rest, and the loss in kinetic energy in this case.
Momentum: \(2mu=2mv_1+mv_2\). Restitution: \(eu=v_2-v_1\).
For \(v_1=0\): \(2u=v_2\), and \(eu=v_2\Rightarrow e=2\), which is impossible (\(e\le1\)). The first sphere cannot be reduced to rest (would require \(e=2\)). Maximum loss occurs at \(e=0\).
Two particles of masses \(m\) and \(3m\) collide directly. The speeds before impact are \(4u\) and \(u\) in opposite directions, and the coefficient of restitution is \(\dfrac{1}{2}\). Find the speeds after impact and the percentage loss in kinetic energy.
Take the direction of the mass \(m\) as positive. Momentum: \(m\cdot4u+3m\cdot(-u)=mv_1+3mv_2\Rightarrow u=v_1+3v_2\).
Restitution: \(\dfrac{1}{2}(4u-(-u))=v_2-v_1\Rightarrow\dfrac{5u}{2}=v_2-v_1\).
Solving: \(v_1=-\dfrac{u}{2}\), \(v_2=\dfrac{u}{2}\). Initial KE \(=\dfrac{1}{2}m\cdot16u^2+\dfrac{1}{2}3m\cdot u^2=9.5mu^2\). Final KE \(=\dfrac{1}{2}m\cdot\dfrac{u^2}{4}+\dfrac{1}{2}3m\cdot\dfrac{u^2}{4}=mu^2\). Loss \(=8.5mu^2\), percentage \(\approx89.5\%\).