SATHEE: Rotation 4 Question 6

Rotation 4 Question 6

7. $A rod A B$ of mass $M$ and length $L$ is lying on a horizontal frictionless surface. A particle of mass $m$ travelling along the surface hits the end $A$ of the rod with a velocity $v_{0}$ in a direction perpendicular to $A B$ . The collision is elastic. After the collision, the particle comes to rest.

(2000)

(a) Find the ratio $m / M$ .

(b) A point $P$ on the rod is at rest immediately after collision. Find the distance $A P$ .

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Answer:

Correct Answer: 7. (a) $\frac{1}{4}$

(b) $\frac{2}{3} L$

Solution:

(a) Let just after collision, velocity of $C M$ of $rod$ is $v$ and angular velocity about $C M$ is $ω$ . Applying following three laws.

(1) External force on the system (rod + mass) in horizontal plane along $x$ -axis is zero. $∴$ Applying conservation of linear momentum in $x$ -direction.

$m v_{0} = m v$

(2) Net torque on the system about CM of rod is zero.

$∴$ Applying conservation of angular momentum about $C M$ of rod, we get $m v_{0} \frac{L}{2} = I ω$

$\begin{aligned} or m v_{0} \frac{L}{2} & = \frac{M L^{2}}{12} ω \\ or m v_{0} & = \frac{M L ω}{6} \end{aligned}$

(3) Since, the collision is elastic, kinetic energy is also conserved.

$\begin{aligned} ∴ & \frac{1}{2} m v_{0}^{2} & = \frac{1}{2} M v^{2} + \frac{1}{2} I ω^{2} \\ or & m v_{0}^{2} & = M v^{2} + \frac{M L^{2}}{12} ω^{2} \end{aligned}$

From Eqs. (i), (ii) and (iii), we get the following results

$\begin{aligned} \frac{m}{M} = \frac{1}{4} \\ v = \frac{m v_{0}}{M} and ω = \frac{6 m v_{0}}{M L} \end{aligned}$

(b) Point $P$ will be at rest if $x ω = v$

or $x = \frac{v}{ω} = \frac{m v_{0} / M}{6 m v_{0} / M L}$

or $x = L / 6$

$t = \frac{π L}{3 v_{0}}$

angle rotated by rod, $θ = ω t = \frac{6 m v_{0}}{M L} \cdot \frac{π L}{3 v_{0}}$

$\begin{aligned} = 2 π \frac{m}{M} \\ = 2 π \frac{1}{4} ∴ θ = \frac{π}{2} \end{aligned}$

Therefore, situation will be as shown below

$∴$ Resultant velocity of point $P$ will be

$\begin{aligned} | v_{P} | & = \sqrt{2} v = \sqrt{2} \frac{m}{M} v_{0} \\ = \frac{\sqrt{2}}{4} v_{0} = \frac{v_{0}}{2 \sqrt{2}} or | v_{P} | = \frac{v_{0}}{2 \sqrt{2}} \end{aligned}$