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

Engineering

Physics

Static and Kinetic Friction

Straight Line Motion

Relative Motion

Question

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

Displacement with respect to ground is zero after 2.66 and displacement with respect to ground is 12 m after 4s

Displacement with respect to ground in 4 s is 4 m.

Displacement with respect to belt in 4 s is −12 m .

Displacement with respect to ground is zero in $\frac{8}{3} s$ .

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

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

So acceleration is 1.5 m/s² 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 × 4 = 16

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

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

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

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} s

so option D is also correct.