Answer:
3.74 %
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 = [tex]\dfrac{1}{2}mv^2 = \dfrac{1}{2}\times 0.1\times v^2[/tex]
Final KE = [tex]\dfrac{1}{2} (M + m) V^2 = \dfrac{1}{2}\times 2.67\times (0.0374 v)^2[/tex]
Now,
the percent of the projectile's KE carried out by target
[tex]=\dfrac{\dfrac{1}{2}\times 2.67\times (0.0374 v)^2}{ \dfrac{1}{2}\times 0.1\times v^2}\times 100[/tex]
[tex]= 3.74\ \%[/tex]
