
- A$4 \sqrt 2 m/s$
- B$4 \sqrt 3 m/s$
- C$4m/s$
- ✓$4 \sqrt 5m/s$

Hence,
change in momentum $=2 \times \rho$
Relative velocity ( ball),
$-v$ along $x-$ axis directed to the right
$-v$ along $y$ -axis directed up
After rebound,
$V$ along $x-$ axis directed to the left
$v$ along $y-$ axis directed down
Therefore,
velocity ( rebounded ball) $=\sqrt{(} 2 v)^{2}+v^{2}$
$v \sqrt{5}$
velocity ( rebounded ball ) $=4 \sqrt{5} \mathrm{m} / \mathrm{s}$
Hence, the velocity of the rebounded ball is $4 \sqrt{5} \mathrm{m} / \mathrm{s}$
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$(A)$ the ring has pure rotation about its stationary $CM$.
$(B)$ the ring comes to a complete stop.
$(C)$ friction between the ring and the ground is to the left.
$(D)$ there is no friction between the ring and the ground.