Unit 4: Linear Momentum
Weightage : 15-25%
In the diagram above, a block of mass \(M\) is initially at rest on a horizontal surface at the base of an inclined plane. The surface and plane have negligible friction. The block is struck by a projectile of mass \(m\) traveling with a horizontal velocity \(v_i\). The projectile becomes embedded in the block, and they move together to the right with speed \(v_f\).
Which of the following is the correct expression for \(v_f\)?
Correct Answer: \( \boxed{\mathrm{C}} \)
Since the projectile becomes embedded in the block, the collision is perfectly inelastic. Therefore, linear momentum is conserved during the collision.
Initial momentum:
\(p_i=mv_i\)
Final momentum:
\(p_f=(m+M)v_f\)
Applying conservation of momentum,
\(mv_i=(m+M)v_f\)
Solving for the final speed,
\(v_f=\dfrac{m}{m+M}v_i\)
The kinetic energy is not conserved because the collision is perfectly inelastic, but momentum is conserved.
Therefore, the correct answer is (C).
A disc of mass \(m\) slides with negligible friction along a flat surface with a velocity \(v\). The disc strikes a wall head-on and bounces back in the opposite direction with a kinetic energy one-fourth of its initial kinetic energy.
What is the final velocity of the disc?
Correct Answer: \( \boxed{\mathrm{D}} \)
The initial and final kinetic energies are related by
\(K_f=\dfrac{1}{4}K_i\)
Using the kinetic energy expression,
\(\dfrac{1}{2}mv_f^2=\dfrac{1}{4}\left(\dfrac{1}{2}mv^2\right)\)
Cancelling the common factors,
\(v_f^2=\dfrac{v^2}{4}\)
Therefore, the magnitude of the final velocity is
\(|v_f|=\dfrac{v}{2}\)
Since the disc rebounds in the opposite direction after striking the wall, the final velocity is negative:
\(v_f=-\dfrac{v}{2}\)
Therefore, the correct answer is (D).
As shown in the top view above, a disc of mass \(m\) is moving horizontally to the right with speed \(v\) on a table with negligible friction when it collides with a second disc of mass \(2m\). The second disc is moving horizontally to the right with speed \( \dfrac{v}{2} \) at the moment of impact. The two discs stick together upon impact.
The speed of the composite body immediately after the collision is
Correct Answer: \( \boxed{\mathrm{C}} \)
Since the discs stick together after the collision, this is a perfectly inelastic collision. Linear momentum is conserved.
Initial momentum:
\(p_{\mathrm{i}} = mv + (2m)\left(\dfrac{v}{2}\right) = mv + mv = 2mv\)
Total mass after collision:
\(M = m + 2m = 3m\)
Applying conservation of linear momentum,
\(2mv = (3m)v_{\mathrm{f}}\)
Therefore,
\(v_{\mathrm{f}} = \dfrac{2mv}{3m} = \dfrac{2v}{3}\)
Thus, the speed of the composite body immediately after the collision is \( \boxed{\dfrac{2v}{3}} \).
Therefore, the correct answer is (C).
