Question

A block is thrown with a velocity of $2m/s$ (relative to ground) on a belt, which is moving with velocity $4m/s$ in opposite direction of the initial velocity of block. If the block stops slipping on the belt after $4s$ of the throwing then choose the correct statement(s)

• A. Displacement with respect to ground is zero after $2.66$ and displacement with respect to ground is $12m$ after $4s$
  
• B. Displacement with respect to ground in $4s$ is $4m$.
• C. Displacement with respect to belt in $4s$ is $-12m$.
• D. Displacement with respect to ground is zero in $\frac{8}{3}s$.

Answer 1

Answer: (B), (C), (D)

The correct options are
B Displacement with respect to ground in $4s$ is $4m$.
C Displacement with respect to belt in $4s$ is $-12m$.
D Displacement with respect to ground is zero in $\frac{8}{3}s$.

Let belt is moving towards left (4m/s) and initial velocity of block is towards right(2m/s).

Now let us assume everthing wrt belt velocity of block will be 6m/s towards right and final velocity will be zero.

So acceleration is $1.5m/s^2$ because time taken to stop wrt belt is 4 sec.

Distance travelled wrt belt is $S=6\times 4-\frac{1}{2}\times 1.5\times 4^2=12$

Distance travelled by belt in that time is $4\times 4=16$

Now taking direction of movement of belt as positive we can say that

Option C. Displacement wrt to belt is -12m.

Option B. Displacement wrt ground is (-12+16=4)=4m.

Also displacement wrt ground will be zero when $S=6\times t-\frac{1}{2}\times 1.5\times t^2-4\times t=0$ $=>t=\frac{8}{3}$ so option D is also correct.