Question

A rifle with a weight of 35 N fires a 4.5-g bullet with a speed of 240 m/s. (a) Find the recoil speed of the rifle in m/s. (b) If a 650-N man holds the rifle firmly against his shoulder, find the recoil speed of the man and rifle in m/s.

Answer 1

The recoil speed of the rifle is 0.0031 m/s when held loosely away from the shoulder. When a 650 N man holds the rifle firmly against his shoulder, the effective mass of the rifle-man system is 28.0 kg, resulting in a recoil speed of 0 m/s.

Step-by-step explanation:

To calculate the recoil speed of the rifle in m/s, we use the principle of conservation of momentum. The momentum of the rifle before firing is equal to the momentum of the bullet after firing. The momentum of an object is calculated by multiplying its mass by its velocity. Given that the mass of the bullet is 4.5 g (0.0045 kg) and the velocity is 240 m/s, we can find the momentum of the bullet. Then, using the principle of conservation of momentum, we can calculate the recoil speed of the rifle.

(a) The momentum of the bullet is calculated as:

Momentum = mass x velocity = 0.0045 kg x 240 m/s = 0.108 kg·m/s

Since the momentum of the bullet before firing is equal to the momentum of the rifle after firing, we can write:

0.108 kg·m/s = mass of the rifle x recoil speed of the rifle

Rearranging the equation, we can solve for the recoil speed of the rifle:

Recoil speed of the rifle = 0.108 kg·m/s ÷ mass of the rifle = 0.108 kg·m/s ÷ 35 N = 0.0030857 m/s

(b) When a 650 N man holds the rifle firmly against his shoulder, the effective mass of the rifle-man system is 28.0 kg. To find the recoil speed of the man and rifle together, we can again use the principle of conservation of momentum. The initial momentum of the rifle-man system is zero, as they are at rest. Therefore, the final momentum of the system after firing must also be zero. We can write:

0 = (mass of the rifle + mass of the man) x recoil speed of the system

Rearranging the equation, we can solve for the recoil speed of the system:

Recoil speed of the system = 0 ÷ (mass of the rifle + mass of the man) = 0 ÷ (28 kg + 650 N ÷ 9.8 m/s²) = 0 m/s

Answer 2

(a) 0.30 m/s

The total momentum of the rifle+bullet system before the shot is zero:

`p_i = 0`

The total momentum of the system after the shot is the sum of the momenta of the rifle and of the bullet:

`p_f = m_r v_r + m_b v_b`

where we have

`m_r = (W)/(g)=(35 N)/(9.8 m/s^2)=3.57 kg` is the mass of the rifle

`v_r` is the final velocity of the rifle

`m_b = 4.5 g = 0.0045 kg` is the mass of the bullet

`v_b = 240 m/s` is the final velocity of the bullet

Since the total momentum must be conserved, we have

`p_i = p_f`

So

`m_r v_r + m_b v_b=0`

and so we can find the recoil velocity of the rifle:

`v_r = - (m_b v_b)/(m_r)=-((0.0045 kg)(240 m/s))/(3.57 kg)=-0.30 m/s`

And the negative sign means it travels in the opposite direction to the bullet: so, the recoil speed is 0.30 m/s.

(b) 0.016 m/s

The mass of the man is equal to its weight divided by the acceleration of gravity:

`m=(W)/(g)=(650 N)/(9.8 m/s^2)=66.3 kg`

This time, we have to consider the system (man+rifle) - bullet. Again, the total momentum of the system before the shot is zero:

`p_i = 0`

while the total momentum after the shot is

`p_f = m_r v_r + m_b v_b`

where this time we have

`m_r = 66.3 kg+3.57 kg=69.9 kg` is the mass of the rifle+person

`v_r` is the final velocity of the man+rifle

`m_b = 4.5 g = 0.0045 kg` is the mass of the bullet

`v_b = 240 m/s` is the final velocity of the bullet

Since the total momentum must be conserved, we have

`m_r v_r + m_b v_b=0`

and so we can find the recoil velocity of the man+rifle:

`v_r = - (m_b v_b)/(m_r)=-((0.0045 kg)(240 m/s))/(66.9 kg)=-0.016 m/s`

So the recoil speed is 0.016 m/s.