Question 361: Centre of Mass and Rotational Motion

Question

A block of mass ' $m$ ' moving on a frictionless surface at speed ' $v$ ' collides elastically with a block of same mass, initially at rest. Now the first block moves at an angle ' $	heta$ ' with its initial direction and has speed ' $v_1$ '. The speed of the second block after collision is

Accepted Answer

The situation can be shown as
Applying law of conservation of kinetic energy,
$$\begin{aligned}
& \mathrm{KE} 	ext { (before collision) }=\mathrm{KE} 	ext { (after collision) } \
& \frac{1}{2} m v^2+\frac{1}{2} m(0)^2=\frac{1}{2} m v_1^2+\frac{1}{2} m v_2^2 \
& \Rightarrow \quad v^2=v_1^2+v_2^2 \Rightarrow v_2=\sqrt{v^2-v_1^2}
\end{aligned}$$
Thus, the velocity of second block after collision is $\sqrt{v^2-v_1^2}$.

Answer

$\sqrt{v_1^2-v^2}$

Answer

$\sqrt{v^2-v_1^2}$

Answer

$\sqrt{v^2+v_1^2}$

Answer

$\sqrt{v-v_1}$

Centre of Mass and Rotational Motion

A block of mass ' m ' moving on a frictionless surface at speed ' v ' collides elastically with a block of same mass, initially at rest. Now the first block moves at an angle ' \theta ' with its initial direction and has speed ' v_1 '. The speed of the second block after collision is

Choose the correct answer: