Question

13.A gun of mass 1000 kg fires a projectile of mass 1 kg with a horizontal velocity of
100 m/s. The velocity of recoil of the gun in the horizontal direction is
(a) 5 m/s
(b) 0.1 m/s (e) 15 m/s (d) 20 m/s​

Answer 1

Answer:

0.1 m/s

Explanation:

hope it'll help you

Answer 2

Answer:

Recoil velocity of the gun is 0.1 m/s in the horizontal direction.

Option b is the correct answer.

Explanation:

The recoil velocity of the gun is solved by using the law of conservation of momentum which states that the total momentum before explosion is equal to the total momentum after explosion. In symbols it is written as $p_1 + p_2 = p_1′ + p_2′.$

Solving for the recoil velocity of the gun $v_1$

$p_1 + p_2 = p_1′ + p_2′$

$m_1v_1 + m_2v_2 = m_1v_1′ + m_2v_2′$

$1000 \: kg \times 0 \: m/s + 1 \:kg \times 0 \:m/s = 1000\: kg \times v_1′ + 1 \: kg \times 100 \:m/s$

$0 + 0 = 1000 \: kg \times v_1′ + 100 \: kg \times \: m/s$

$1000 \: kg \times ( -v_1′ )= 100 \: kg \times m/s$

$-v_1′ = \frac{100 \: kg \times m/s}{1000 \: kg}$

$-v_1′ = 0.1 \:m/s$

$v_1′ = -0.1 \:m/s$…….. the negative sign means that the recoil velocity is opposite in direction to that of the bullet.

The recoil velocity of the gun is -0.1 m/s