A 6 kg rifle is used to fire a 9 g bullet, which travels with a speed of 700 m/s.
(a) What is the speed of recoil of the rifle?
______m/s
(b) How much energy does it transmit to the shoulder of the person using the rifle as it stops?
______J
I have solved part (a) which is 1.05 m/s but I have no idea how to do part (b)
To calculate the energy transmitted to the shoulder of the person using the rifle as it stops (part b), we need to consider the principle of conservation of momentum and kinetic energy.
The key observation is that the total momentum before the bullet is fired is zero, and thus, the total momentum after the bullet is fired should also be zero. This is because there is no external force acting on the system (rifle + bullet) in the horizontal direction.
The momentum of an object is given by the product of its mass and velocity. Let's assume the initial velocity of the rifle is v, and the final velocity of the rifle (after firing the bullet) is v'.
According to the law of conservation of momentum:
M_rifle * v = m_bullet * v' -- (1)
where M_rifle is the mass of the rifle and m_bullet is the mass of the bullet.
In this case, the initial momentum is zero as there is no initial velocity for both the rifle and the bullet (since they are at rest). Thus, we can rewrite equation (1) as:
0 = m_bullet * v' -- (2)
Furthermore, the kinetic energy of an object is given by the formula:
KE = (1/2) * mass * velocity^2
The initial kinetic energy of the rifle (before firing) is zero since it is at rest. The kinetic energy of the bullet can be calculated as:
KE_bullet = (1/2) * m_bullet * v_bullet^2
where v_bullet is the velocity of the bullet.
According to the law of conservation of energy, the total initial kinetic energy should be equal to the total final kinetic energy.
0 + KE_bullet = KE_rifle + 0
Simplifying, we find:
(1/2) * m_bullet * v_bullet^2 = (1/2) * M_rifle * v'^2
Since v' is the velocity of the recoil of the rifle, the final velocity of the rifle after firing the bullet, we can use the equation (2) to rewrite the above equation as:
(1/2) * m_bullet * v_bullet^2 = (1/2) * M_rifle * (0^2)
Simplifying further,
m_bullet * v_bullet^2 = 0
Which ultimately implies that v_bullet = 0, as the mass of the bullet and the velocity of the bullet cannot be zero simultaneously.
Therefore, there is no kinetic energy transfer to the rifle. Hence, there is no energy transmitted to the shoulder of the person using the rifle as it stops, i.e., the answer to part (b) is 0 J.