A pendulum consists of a wooden bob of mass m and of length l. A bullet of mass m1 is fired towards the pendulum with a speed v1 The bullet emerges out of the bob with a speed v1/3 and the bob just completes motion along a vertical circle. Then v1 is:

Question

A pendulum consists of a wooden bob of mass m and of length l. A bullet of mass m1 is fired towards the pendulum with a speed v1 The bullet emerges out of the bob with a speed v1/3 and the bob just completes motion along a vertical circle. Then v1 is: (A) ((m/m1))√5gl (B) (3/2)((m/m1))√5gl (C) (2/3)((m1/m2))√5gl (D) ((m1/m))√gl

Tardigrade Admin · Accepted Answer

Applying the law of conservation of momentum m1 v1=m1 (v1/3)+m v or v=(2 m1 dotv1/3 m) To describe a vertical circle v should be √5gl So (2 m1 v1/3 m)=√5 g l or v1=((m/m1)) (3/2) √5 g l

Q. A pendulum consists of a wooden bob of mass $m$ and of length $l . A$ bullet of mass $m_{1}$ is fired towards the pendulum with a speed $v_{1}$ The bullet emerges out of the bob with a speed $v_{1} /3$ and the bob just completes motion along a vertical circle. Then $v_{1}$ is:

$(\frac{m}{m _{1}}) 5 g l$

$\frac{3}{2} (\frac{m}{m _{1}}) 5 g l$

$\frac{2}{3} (\frac{m _{1}}{m _{2}}) 5 g l$

$(\frac{m _{1}}{m}) g l$

Applying the law of conservation of momentum
$m_{1} v_{1} = m_{1} \frac{v _{1}}{3} + m v$
or $v = \frac{2 m _{1} v ˙ _{1}}{3 m}$
To describe a vertical circle v should be $5 g l$
So $\frac{2 m _{1} v _{1}}{3 m} = 5 g l$
or $v_{1} = (\frac{m}{m _{1}}) \frac{3}{2} 5 g l$

Q. A pendulum consists of a wooden bob of mass m and of length l.A bullet of mass m1​ is fired towards the pendulum with a speed v1​ The bullet emerges out of the bob with a speed v1​/3 and the bob just completes motion along a vertical circle. Then v1​ is:

(m1​m​)5gl​

23​(m1​m​)5gl​

32​(m2​m1​​)5gl​

(mm1​​)gl​