Question

An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are placed very close to a pendulum bob of mass 10.0 kg attached to a massless rod of length L. Assume that catapult 1 fires a projectile of mass 5.50 g at speed v1 and catapult 2 fires a projectile of mass 10.0 g at speed v2. If the projectile from catapult 1 causes the pendulum to swing to a maximum angular displacement of 6.80 ∘ and the projectile from catapult 2 causes a displacement of 11.4 ∘, find the ratio of the initial speeds, v1v2 .

Answer 1

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

$mu=(M+m)v$

$v=\frac{m}{M+m}\times u$

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

$\frac{(M+m)v^2}{2}=(M+m)gh$

here $h=L(1-\cos \theta )$ from diagram

therefore

$v=\sqrt{2gL(1-\cos \theta )}$

initial velocity in terms of v

$u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}$

For first case $\theta =6.8^{\circ}$

$u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}$

for second case $\theta =11.4^{\circ}$

$u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}$

Therefore $\frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}$

$\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}$

$\frac{u_1}{u_2}=1.084$

i.e.$\frac{v_1}{v_2}=1.084$