Two small balls A and B, each of mass m, are joined rigidly to the ends of a light rod of lengh L (figure). The system translates on a frictionless horizontal surface with a velocity vo in a direction perpendicular to the rod. A particle P of mass m kept at rest on the surface sticks to the ball A as the ball collides with it. Find:
- The linear speeds of the balls A and B after the collision.
- The velocity of the centre of mass C of the system A + B + P.
- The angular speed of the system about C after the collision.
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Two balls A & B, each of mass m are joined rigidly to the ends of a light of rod of length L. The system moves in a velocity $v_0$ in a direction $\perp$ to the rod. A particle P of mass m kept at rest on the surface sticks to the ball A as the ball collides with it.
If we consider the three bodies to be a system.
Applying L.C.L.M.
Therefore $\text{mv}_0=2\text{mv}'$
$\Rightarrow\text{v}'=\frac{\text{v}_0}{2}$
Therefore a has velocity $\frac{\text{v}_0}{2}$
Therefore $\text{V}_{\text{cm}}=\frac{\text{m}\times\text{v}_0+2\text{m}\big(\frac{\text{v}_0}{2}\big)}{\text{m}+\text{2m}}$
$=\frac{\text{mv}_0+\text{mv}_0}{3\text{m}}=\frac{2\text{v}_0}{3}$ (along the initial velocity as before collision)
[Only magnitude has been taken]
Distance of the (A + P) from centre of mass $=\frac{\text{l}}3{}$ & for B it is $=\frac{2\text{l}}3{}.$
Therefore $\text{P}_\text{cm}=\text{l}_\text{cm}\times\omega$
$\Rightarrow2\text{m}\times\frac{\text{v}_0}{6}\times\frac{1}{3}+\text{m}\times\frac{\text{v}_0}{3}\times\frac{\text{2l}}{3}$
$=\text{2m}\Big(\frac{1}{3}\Big)^2+\text{m}\Big(\frac{2\text{l}}{3}\Big)^2\times\omega$
$\Rightarrow\frac{6\text{mv}_0\text{l}}{18}=\frac{\text{6ml}}{9}\times\omega$
$\Rightarrow\omega=\frac{\text{v}_0}{\text{2l}}$
- The light rod will exert a force on the ball B only along its length. So collision will not affect its velocity.
If we consider the three bodies to be a system.
Applying L.C.L.M.
Therefore $\text{mv}_0=2\text{mv}'$
$\Rightarrow\text{v}'=\frac{\text{v}_0}{2}$
Therefore a has velocity $\frac{\text{v}_0}{2}$
- if we consider the three bodies to be a system.
Therefore $\text{V}_{\text{cm}}=\frac{\text{m}\times\text{v}_0+2\text{m}\big(\frac{\text{v}_0}{2}\big)}{\text{m}+\text{2m}}$
$=\frac{\text{mv}_0+\text{mv}_0}{3\text{m}}=\frac{2\text{v}_0}{3}$ (along the initial velocity as before collision)
- The velocity of (A + P) w.r.t. the centre of mass $=\frac{2\text{v}_0}{3}-\frac{\text{v}_0}{2}=\frac{\text{v}_0}{6}$ &
[Only magnitude has been taken]
Distance of the (A + P) from centre of mass $=\frac{\text{l}}3{}$ & for B it is $=\frac{2\text{l}}3{}.$
Therefore $\text{P}_\text{cm}=\text{l}_\text{cm}\times\omega$
$\Rightarrow2\text{m}\times\frac{\text{v}_0}{6}\times\frac{1}{3}+\text{m}\times\frac{\text{v}_0}{3}\times\frac{\text{2l}}{3}$
$=\text{2m}\Big(\frac{1}{3}\Big)^2+\text{m}\Big(\frac{2\text{l}}{3}\Big)^2\times\omega$
$\Rightarrow\frac{6\text{mv}_0\text{l}}{18}=\frac{\text{6ml}}{9}\times\omega$
$\Rightarrow\omega=\frac{\text{v}_0}{\text{2l}}$
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