Question

man fires A bullet of mass 200 g at a speed of 5 metre per second . the gun is of 1kg by what velocity of the gun reBond​

Answer 1

Answer :-

-1m/s

Explanation :-

Given :

Mass of the bullet,m_1 = 200g = 0.2kg

Initial velocity of the bullet,u_1 = 0    [as bullet was in rest before firing]

Final velocity of the bullet,v_1 = 5m/s

Mass of the gun,m_2 = 1kg

Initial velocity of the gun,u_1 = 0    [as gun was in rest before firing]

To Find :

Rebound velocity of the gun,v = ?

Solution :

According to the law of covervation of momentum,

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

Put their values and find “v_2”

$\sf{}:\implies 0.2kg \times 0m/s+1kg\times 0m/s=0.2kg\times 5m/s+1kg \times v_2$

$\sf{}:\implies 0=1.0kg\ m/s+1kg \times v_2$

$\sf{}:\implies -1kg\ m/s=+1kg \times v_2$

$\sf{}:\implies -\dfrac{1kg\ m/s}{1kg}=v_2$

$\sf{}\thereofre v_2=-1m/s$

Therefore,rebound velocity of the gun is equal to -1m/s