Home / CIE AS & A Level Physics : 3.3 Linear momentum and its conservation – Exam style question – Paper 1
An object of mass \(2.0\,\mathrm{kg}\) is travelling at a speed of \(3.0\,\mathrm{m\,s^{-1}}\) on a horizontal frictionless surface. This object collides head-on with a stationary object of mass \(1.0\,\mathrm{kg}\). The two objects stick together on impact.
How much kinetic energy is lost on impact?
Correct Answer: \( \boxed{\mathrm{D}} \)
Initial kinetic energy \(=\dfrac{1}{2}(2.0)(3.0)^2=9.0\,\mathrm{J}\).
By conservation of momentum, the final speed is \(v=\dfrac{2.0\times3.0}{3.0}=2.0\,\mathrm{m\,s^{-1}}\).
Final kinetic energy \(=\dfrac{1}{2}(3.0)(2.0)^2=6.0\,\mathrm{J}\).
Kinetic energy lost \(=9.0-6.0=3.0\,\mathrm{J}\).
Therefore, the correct answer is (D).
Two different blocks, P and Q, slide towards each other on a horizontal frictionless surface. The blocks have an elastic collision.
The diagram shows the velocities of the two blocks immediately after the collision.
Which row gives possible velocities of the two blocks immediately before the collision?
| Velocity of P | Velocity of Q | |
|---|---|---|
| (A) | \(10\,\mathrm{cm\,s^{-1}}\) to the right | \(60\,\mathrm{cm\,s^{-1}}\) to the left |
| (B) | \(50\,\mathrm{cm\,s^{-1}}\) to the right | zero |
| (C) | \(20\,\mathrm{cm\,s^{-1}}\) to the left | \(70\,\mathrm{cm\,s^{-1}}\) to the right |
| (D) | \(60\,\mathrm{cm\,s^{-1}}\) to the right | \(30\,\mathrm{cm\,s^{-1}}\) to the left |
Correct Answer: \( \boxed{\mathrm{D}} \)
In an elastic collision, both momentum and kinetic energy are conserved.
The relative speed of approach equals the relative speed of separation.
Only option (D) is consistent with these conservation laws and the given velocities after the collision.
Therefore, the correct answer is (D).
Two balls, of masses \(m\) and \(2m\), travelling in a vacuum with initial velocities \(2v\) and \(v\) respectively, collide with each other head-on, as shown.
After the collision, the ball of mass \(m\) rebounds to the left with velocity \(v\).
What is the loss of kinetic energy in the collision?
Correct Answer: \( \boxed{\mathrm{C}} \)
Take motion to the right as positive.
Initial momentum:
\(m(2v)+2m(-v)=0\)
Hence the final momentum is also zero.
After the collision, the ball of mass \(m\) has velocity \(-v\). Let the velocity of the \(2m\) ball be \(u\).
\(-mv+2mu=0 \Rightarrow u=\dfrac{v}{2}\)
Initial kinetic energy:
\(KE_i=\dfrac{1}{2}m(2v)^2+\dfrac{1}{2}(2m)v^2=3mv^2\)
Final kinetic energy:
\(KE_f=\dfrac{1}{2}mv^2+\dfrac{1}{2}(2m)\left(\dfrac{v}{2}\right)^2=\dfrac{3}{4}mv^2\)
Loss of kinetic energy:
\(KE_i-KE_f=3mv^2-\dfrac{3}{4}mv^2=\dfrac{9}{4}mv^2\)
Therefore, the correct answer is (C).