A 7.45-g bullet is moving horizontally with a velocity of +348 m/s, where the sign + indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1220 g, and its velocity is +0.648 m/s after the bullet passes through it. The mass of the second block is 1603 g. (a) What is the velocity of the second block after the bullet imbeds itself? (b) Find the ratio of the total kinetic energy after the collision to that before the collision.

Answer 1

m=7,45 g, m1=1220 g, m2=1603 g,
v=348 m/s, v1 = 0.648 m/s
The law of conservation of linear momentum for the 1st collision
m•v+0=m1•v1+m•u
u=(m•v-m1•v1)/m = v –(m1/m)v1=
348 – (1220/7.45) •0.648 = 241.9 m/s,
The law of conservation of linear momentum for the 2nd collision
mu+0=(m+m1)u1
u1= mu/(m1+m) = 7.45•241.9/(1603+7.45) =1.12 m/s.
Before the 1st collision KE= m•v²/2,
After the 1st collision
KE1=m1•v1²/2+m•u²/2
After the 2nd collision
KE2=(m+m1) •u1²/2