Question

A projectile with a mass of 0.100 kg is fired at and embeds itself in a stationary target with a mass of 2.57 kg. The target and embedded projectile fly off after being struck. Determine the percent of the projectile's incident kinetic energy carried off by the target-projectile combination.

Answer 1

3.74 %

Step-by-step explanation:

Given,

Mass of the fired projectile, m = 0.1 Kg

Mass of the stationary target, M = 2.57 Kg

Speed of the particle after collision

Using conservation of momentum

m v = (m + M) V

0.1 x v = (0.1 + 2.57) V

V = 0.0374 v

Initial KE =
`(1)/(2)mv^2 = (1)/(2)* 0.1* v^2`

Final KE =
`(1)/(2) (M + m) V^2 = (1)/(2)* 2.67* (0.0374 v)^2`

Now,

the percent of the projectile's KE carried out by target

`=((1)/(2)* 2.67* (0.0374 v)^2)/( (1)/(2)* 0.1* v^2)* 100`

`= 3.74\ \%`