Question

Calculate the recoil velocity in the horizontal direction, in meters per second, of a 1.25-kg plunger that directly interacts with a 0.0175-kg bullet fired at 585 m/s from the gun. Take the firing direction to be the positive direction.

Answer 1

Answer:

$v_1=-8.19\ m/s$'

Explanation:

It is given that,

Mass of the plunger, $m_1=1.25\ kg$

Mass of the bullet, $m_2=0.0175\ kg$

Initially both plunger and the bullet are at rest, $u_1=u_2=0$

Final speed of the bullet, $v_2=585\ m/s$

Let $v_1$ is the final speed of the plunger. Using the conservation of momentum to find it. The equation is as follows :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

Since, $u_1=u_2=0$

$m_1v_1+m_2v_2=0$

$v_1=-\dfrac{m_2v_2}{m_1}$

$v_1=-\dfrac{0.0175\times 585}{1.25}$

$v_1=-8.19\ m/s$

So, the recoil velocity of the plunger is 8.19 m/s. Hence, this is the required solution.