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Question

A block is thrown with a velocity of 2 ms–1 (relative to ground) on a belt, which is moving with velocity 4 ms–1 in the opposite direction of the initial velocity of block. If the block stops slipping on the belt after 4 s after it is thrown, then choose the correct statements(s). Assume constant deceleration of the block.

A
Displacement with respect to ground is zero after 2.66 s and magnitude of displacement with respect to ground is 12 m after 4 s.
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B
Magnitude of displacement with respect to ground in 4 s is 4 m.
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C
Magnitude of displacement with respect to belt in 4 s is 12 m.
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D
Displacement with respect to ground is zero in 8/3 s.
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Solution

The correct options are
B Magnitude of displacement with respect to ground in 4 s is 4 m.
C Magnitude of displacement with respect to belt in 4 s is 12 m.
D Displacement with respect to ground is zero in 8/3 s.
v=u+at
4=2+a×4
a=32 m/s2
(B) Now s=2×412×32×(4)2
=812=4, (with respect to ground)
(C) Relative velocity ut=6 m/s and v=0
s=6×412×32×(4)2
=2412=12 m, (with respect to belt)
(D) Displacement with respect to ground is zero
0=2×t12×32×t2
t=83 s

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