A uniform rod of length L and mass M lies on a smooth horizontal...

A uniform rod of length

L

and mass

M

lies on a smooth horizontal table. A particle of mass

m

moving on the table strikes the rod perpendicularly at one end with speed

v

and sticks to it. (a) Find the velocity of the centre of mass

C

of the system "rod + particle". (b) Find the velocity of the particle with respect to

C

before the collision. (c) Find the velocity of the rod with respect to

C

before the collision. (d) Find the angular momentum of the particle and of the rod about

C

before the collision. (e) Find the moment of inertia of the system about the vertical axis through

C

after the collision. (f) Find the velocity of

C

and the angular velocity of the system about

C

after the collision.

Options

A(a) $\dfrac{mv}{M + m}$ (b) $\dfrac{Mv}{M + m}$ (c) $-\dfrac{mv}{M + m}$ (d) $\dfrac{M^2 m v L}{2(M + m)^2}$ , $\dfrac{M m^2 v L}{2(M + m)^2}$ (e) $\dfrac{M(M + 4m)L^2}{12(M + m)}$ (f) $\dfrac{mv}{M + m}$ , $\dfrac{6mv}{(M + 4m)L}$
B(a) $\dfrac{mv}{M + m}$ (b) $\dfrac{Mv}{M + m}$ (c) $-\dfrac{mv}{M + m}$ (d) $\dfrac{M^2 m v L}{(M + m)^2}$ , $\dfrac{M m^2 v L}{(M + m)^2}$ (e) $\dfrac{M(M + 2m)L^2}{12(M + m)}$ (f) $\dfrac{mv}{M + m}$ , $\dfrac{3mv}{(M + 2m)L}$
C(a) $\dfrac{mv}{M + m}$ (b) $\dfrac{mv}{M + m}$ (c) $-\dfrac{Mv}{M + m}$ (d) $\dfrac{M m v L}{(M + m)^2}$ , $\dfrac{M m v L}{(M + m)^2}$ (e) $\dfrac{(M + m)L^2}{12}$ (f) $\dfrac{mv}{M + m}$ , $\dfrac{6mv}{(M + m)L}$
D(a) $\dfrac{mv}{M}$ (b) $v$ (c) $0$ (d) $0$ , $0$ (e) $\dfrac{ML^2}{12}$ (f) $\dfrac{mv}{M}$ , $0$

Correct answer

A. (a) $\dfrac{mv}{M + m}$ (b) $\dfrac{Mv}{M + m}$ (c) $-\dfrac{mv}{M + m}$ (d) $\dfrac{M^2 m v L}{2(M + m)^2}$ , $\dfrac{M m^2 v L}{2(M + m)^2}$ (e) $\dfrac{M(M + 4m)L^2}{12(M + m)}$ (f) $\dfrac{mv}{M + m}$ , $\dfrac{6mv}{(M + 4m)L}$

Step-by-step solution

(a)

\dfrac{mv}{M + m}

(b)

\dfrac{Mv}{M + m}

(c)

-\dfrac{mv}{M + m}

(d)

\dfrac{M^2 mvl}{2(M + m)^2}

\dfrac{Mm^2 vl}{2(M + m)^2}

(e)

\dfrac{M(M + 4m)L^2}{12(M + m)}

(f)

\dfrac{mv}{M + m}

\dfrac{6mv}{(M + 4m)L}

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