Question

Two point masses A and B each of 1 Kg joined by 1 m rod (light) translates on a horizontal surface as shown in the figure. A particle of mass 1 Kg 1 Kg at rest on horizontal surface sticks to ball as it collides with ball A. Then

• A. Linear speed of a ball A just after collision is 2 m/s
• B. Linear speed of a ball B just after collision is 4 m/s
• C. Velocity of cenetr of mass of system is 8/3 m/s
• D. Angular speed of system about center of mass after collision 2nd/s

Answer 1

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

The correct options are
A Linear speed of a ball A just after collision is 2 m/s
B Linear speed of a ball B just after collision is 4 m/s
C Velocity of cenetr of mass of system is 8/3 m/s
D Angular speed of system about center of mass after collision 2nd/s

Let be the velocity of particle B $v_B=4m/s$ and that of particle A is $v_A=4m/s$.

By looking at the given figure we can say that the light rod will exert a force on the ball B only along its length. So collision will not effect its velocity.

Particle B will have a velocity =$v_B=4m/s$

Velocity of particle A when we consider the three bodies to be a system:

$m{v}_B=2m{v}_A$

$v_A=\frac{v_B}{2}$

$v_A=\frac{4}{2}$

$v_A=2m/s$

The velocity of the centre of mass of the system is:

$v_{cm}=\frac{m{v}_B+2m\left(\frac{v_b}{2}\right)}{m+2m}$

$v_{cm}=\frac{m{v}_B+m{v}_B}{3m}$

$v_{cm}=\frac{2v_B}{3}$

$v_{cm}=\frac{8}{3}m/s$

The velocity of A with respect to the centre of mass:

$\frac{2v_B}{3}-\frac{v_B}{2}=\frac{v_B}{6}$

The velocity of B with respect to the centre of mass:

$v_B-\frac{2v_B}{3}=\frac{v_B}{3}$

Distance of the A from centre of mass =$\frac{1}{3}$ and for B =$\frac{2l}{3}$

Therfore;

$P_{cm}=I_{cm}\times \omega$

$2m\times \frac{v_B}{6}\times \frac{1}{3}+m\times \frac{v_B}{3}\times \frac{2l}{3}$

$2m(\frac{1}{3}{)}^2+m(\frac{2l}{3}{)}^2\times \omega$

$\frac{6m{v}_{B}l}{18}=\frac{6ml}{9}\times \omega$

$\omega =\frac{v_B}{2l}$

$\omega =\frac{4}{2}$

$\omega =2rad/sec$

All the above mentioned options are correct.