Question

A projectile (mass = 0.25 kg) is fired at and embeds itself in a stationary target (mass = 2.54 kg). With what percentage of the projectile's incident kinetic energy does the target (with the projectile in it) fly off after being struck?

Answer 1

By collision theory,

" The total momentum of the bodies before the collision must be equal to the total momentum after the collision"

We know that

Momentum = Mass x Velocity

Thus by this,

Total momentum before the collision = momentum of the projectile + momentum of the target.

Before Collision:

The momentum of the projectile, P_p = mass of projectile x velocity before collision

P_p = m_p x V_p

  Where

m_p = mass of the projectile

V_p = velocity of the projectile before the collision

Kinetic Energy of the projectile before impact, KE_p = 0.5 x m_p x V_p²

The momentum of the target before the collision, P_t = 0 as the target is stationary.

After the collision:

The total mass of the body = mass of the projectile + mass of the target = m_p + m_t

Total momentum after collision, P_t+p = ( m_p + m_t ) x V_t+p

where

V_t+p = final velocity of the embedded target + projectile.

The Kinetic Energy of the target after the collision, KE_t+p = 0.5 x ( m_p + m_t ) x V²_t+p

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Equating the conditions before and after the collision,

P_p = P_t+p

or,   

or,   

This is the velocity of the target after collision

Thus,

percentage of the projectile's incident kinetic energy that the target flies off,

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