Question

A 5.17-g bullet is moving horizontally with a velocity of +369 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 1213 g, and its velocity is +0.566 m/s after the bullet passes through it. The mass of the second block is 1501 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

mass of bullet mb = 5.17 g

mass of block 1 , m1 = 1213 g

mass of second block, m2 = 1501 g

initial speed of bullet before collision vb0 = 369 m/s

during collision with m1

momentum before 1st collision pi = mb*vb0

momentum after collision Pf = mb*vb1 + m1*v1

from conservation of momentum Pf = Pi

mb*vb1 + m1*v1 = mb*vb0

5.17*vb1 + 1213*0.566 = 5.17*369

vb1 = 236.2 m/s

speed of bullet after passing the block1 = 236.2 m/s

during collision with second block ( inelastic collision)

momentum before second collision pi = mb*vb1

momentum after second collision pf = (mb+m2)*v

from momentum conservation Pf = Pi

(5.17 + 1501)*v = 5.17*236.2

v = 0.811 m/s <<<----------ANSWER

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part B

kinetic energy before collision Ki = (1/2)*mb*vb0^2 = (1/2)*5.17*10^-3*369^2

kinetic energy before collision Ki = 352 J

kinetic energy after collision Kf = (1/2)*mb*v^2 + (1/2)*m1*v1^2 + (1/2)*m2*v^2

Kf = (1/2)*5.17*10^-3*0.811^2 + (1/2)*1213*10^-3*0.566^2+(1/2)*1501*10^-3*0.811^2

Kf = 0.689 J

ration = Kf/Ki = 2*10^-3