Question

A block is thrown with a velocity of $2{\text{ms}}^{–1}$ (relative to ground) on a belt, which is moving with velocity $4{\text{ms}}^{–1}$ in the opposite direction of the initial velocity of block. If the block stops slipping on the belt after $4\text{s}$ 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 $2.66\text{s}$ and magnitude of displacement with respect to ground is $12\text{m}$ after $4\text{s}$.
• B. Magnitude of displacement with respect to ground in $4\text{s}$ is $4\text{m}$.
• C. Magnitude of displacement with respect to belt in $4\text{s}$ is $12\text{m}$.
• D. Displacement with respect to ground is zero in 8/3 s.

Answer 1

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

The correct options are
B Magnitude of displacement with respect to ground in $4\text{s}$ is $4\text{m}$.
C Magnitude of displacement with respect to belt in $4\text{s}$ is $12\text{m}$.
D Displacement with respect to ground is zero in 8/3 s.

$v=u+at$
$\therefore -4=2+a\times 4$
$a=-\frac{3}{2}{\text{m/s}}^2$
(B) Now $s=2\times 4-\frac{1}{2}\times \frac{3}{2}\times (4{)}^2$
$=8-12=-4$, (with respect to ground)
(C) Relative velocity $u_t=6\text{m/s}$ and $v=0$
$s=6\times 4-\frac{1}{2}\times \frac{3}{2}\times (4{)}^2$
$=24-12=12\text{m}$, (with respect to belt)
(D) Displacement with respect to ground is zero
$0=2\times t-\frac{1}{2}\times \frac{3}{2}\times t^2$
$t=\frac{8}{3}\text{s}$