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Question: Two blocks $A$ and $B$, each of mass $m$, are connected by an ideal spring of stiffness $k$ and plac...

Two blocks AA and BB, each of mass mm, are connected by an ideal spring of stiffness kk and placed on a smooth horizontal surface. A ball of mass mm moving with a velocity v0v_0 strikes the block AA and gets embedded to it. Then

A

Velocity of block AA just after collision is v02\frac{v_0}{2}

B

Velocity of block BB just after collision is zero

C

The maximum compression produced in the spring is v0m6kv_0\sqrt{\frac{m}{6k}}

D

The kinetic energy lost during collision is 14mv02\frac{1}{4}mv_0^2

Answer

A, B, C, D

Explanation

Solution

The problem involves two stages: an inelastic collision followed by the oscillatory motion of a spring-mass system.

Stage 1: Collision between the ball and block A

A ball of mass mm moving with velocity v0v_0 strikes block A (mass mm) and gets embedded in it. This is a perfectly inelastic collision.
Let vA′v_A' be the velocity of the combined mass (ball + block A) just after the collision.
Applying the principle of conservation of linear momentum:
Initial momentum = Final momentum
mv0+m(0)=(m+m)vA′m v_0 + m (0) = (m+m) v_A'
mv0=2mvA′m v_0 = 2m v_A'
vA′=v02v_A' = \frac{v_0}{2}

  • Option (A) Velocity of block A just after collision is v02\frac{v_0}{2}
    This is correct, as calculated above.

  • Option (B) Velocity of block B just after collision is zero
    Block B is not directly involved in the collision and is connected by a spring. The spring takes time to compress or expand and transmit force. Therefore, instantaneously, just after the collision, block B's velocity remains unchanged.
    This is correct.

Stage 2: Motion of the combined system (ball+A) and block B

After the collision, the combined mass (ball + block A), let's call it M1=2mM_1 = 2m, moves with velocity vA′=v02v_A' = \frac{v_0}{2}. Block B, let's call it M2=mM_2 = m, is at rest (vB=0v_B = 0). The spring is initially at its natural length.
The maximum compression in the spring occurs when both masses move with a common velocity, say vfv_f. At this point, the relative velocity between M1M_1 and M2M_2 is zero.

Applying the principle of conservation of linear momentum for the system (M1+M2M_1 + M_2):
Initial momentum (just after collision) = Final momentum (at maximum compression)
M1vA′+M2vB=(M1+M2)vfM_1 v_A' + M_2 v_B = (M_1 + M_2) v_f
(2m)(v02)+m(0)=(2m+m)vf(2m) \left(\frac{v_0}{2}\right) + m (0) = (2m+m) v_f
mv0=3mvfm v_0 = 3m v_f
vf=v03v_f = \frac{v_0}{3}

Now, apply the principle of conservation of mechanical energy from just after the collision to the point of maximum compression.
Initial mechanical energy (just after collision):
KEinitial=12M1(vA′)2+12M2(vB)2=12(2m)(v02)2+12m(0)2=12(2m)v024=mv024KE_{initial} = \frac{1}{2} M_1 (v_A')^2 + \frac{1}{2} M_2 (v_B)^2 = \frac{1}{2} (2m) \left(\frac{v_0}{2}\right)^2 + \frac{1}{2} m (0)^2 = \frac{1}{2} (2m) \frac{v_0^2}{4} = \frac{m v_0^2}{4}
PEinitial=0PE_{initial} = 0 (spring is at natural length)

Final mechanical energy (at maximum compression xmaxx_{max}):
KEfinal=12(M1+M2)vf2=12(2m+m)(v03)2=12(3m)v029=mv026KE_{final} = \frac{1}{2} (M_1 + M_2) v_f^2 = \frac{1}{2} (2m+m) \left(\frac{v_0}{3}\right)^2 = \frac{1}{2} (3m) \frac{v_0^2}{9} = \frac{m v_0^2}{6}
PEfinal=12kxmax2PE_{final} = \frac{1}{2} k x_{max}^2

By conservation of mechanical energy: KEinitial+PEinitial=KEfinal+PEfinalKE_{initial} + PE_{initial} = KE_{final} + PE_{final}
mv024+0=mv026+12kxmax2\frac{m v_0^2}{4} + 0 = \frac{m v_0^2}{6} + \frac{1}{2} k x_{max}^2
12kxmax2=mv024−mv026\frac{1}{2} k x_{max}^2 = \frac{m v_0^2}{4} - \frac{m v_0^2}{6}
12kxmax2=mv02(3−212)\frac{1}{2} k x_{max}^2 = m v_0^2 \left(\frac{3-2}{12}\right)
12kxmax2=mv0212\frac{1}{2} k x_{max}^2 = \frac{m v_0^2}{12}
kxmax2=mv026k x_{max}^2 = \frac{m v_0^2}{6}
xmax2=mv026kx_{max}^2 = \frac{m v_0^2}{6k}
xmax=v0m6kx_{max} = v_0 \sqrt{\frac{m}{6k}}

  • Option (C) The maximum compression produced in the spring is v0m6kv_0\sqrt{\frac{m}{6k}}
    This is correct, as calculated above.

Stage 3: Kinetic energy lost during collision

The kinetic energy lost occurs only during the inelastic collision between the ball and block A.
Initial kinetic energy (before collision):
KEbefore=12mv02+12m(0)2=12mv02KE_{before} = \frac{1}{2} m v_0^2 + \frac{1}{2} m (0)^2 = \frac{1}{2} m v_0^2
Final kinetic energy (just after collision):
KEafter=12(2m)(vA′)2=12(2m)(v02)2=12(2m)v024=mv024KE_{after} = \frac{1}{2} (2m) (v_A')^2 = \frac{1}{2} (2m) \left(\frac{v_0}{2}\right)^2 = \frac{1}{2} (2m) \frac{v_0^2}{4} = \frac{m v_0^2}{4}

Kinetic energy lost = KEbefore−KEafterKE_{before} - KE_{after}
KElost=12mv02−mv024=(12−14)mv02=14mv02KE_{lost} = \frac{1}{2} m v_0^2 - \frac{m v_0^2}{4} = \left(\frac{1}{2} - \frac{1}{4}\right) m v_0^2 = \frac{1}{4} m v_0^2

  • Option (D) The kinetic energy lost during collision is 14mv02\frac{1}{4}mv_0^2
    This is correct, as calculated above.

All four options are correct.