Question

A bullet of 5gm is fired from a pistol of 1.5 kg.If the recoil velocity of the pistol is 1.5m/s.Find the velocity of the bullet.

Answer 1

Answer:-

-450m/s

Explanation:-

Given:

$\sf{}Mass\ of\ the\ bullet(m_1)= 5gm= 0.005kg$

$\sf{}Mass\ of\ the\ pistol(m_2)= 1.5kg$

$\sf{}Recoil\ velocity\ of\ pistol(v_2)=1.5m/s$

Since,before firing the bullet and pistol were in rest

$\sf{}Initial\ velocity\ of\ bullet(u_1)=0$

$\sf{}Initial\ velocity\ of\ pistol(u_2)=0$

To Find:

$\sf{}Recoil\ velocity\ of\ bullet(v_1)= ?$

Solution:

According to conversation law of momentum

$\sf{}m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$\sf{}\Rightarrow 0.005kg\times 0+1.5kg\times 0=0.005v_1+1.5kg\times1.5m/s$

$\sf{}\Rightarrow 0=0.005v_1+2.25kg\ m/s$

$\sf{}\Rightarrow -2.25kg=0.005kg\timesv_1\ m/s$

$\sf{}\Rightarrow -\dfrac{2.25kg}{0.005}=v_1\ m/s$

$\sf{}\Rightarrow v_1 = -450m/s$

Thus,velocity of the bullet = -450m/s

Here negative sign with a velocity of the pistol shows that,bullet moves in the opposite direction of the pistol.

Answer 2

Answer:

hi

Explanation: