Question

A 4.80-g bullet is moving horizontally with a velocity of +357 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 1150 g, and its velocity is +0.602 m/s after the bullet passes through it. The mass of the second block is 1529 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

Solution:

Mass of bullet = m = 4.8 g

Horizontal velocity = 357 m/s

Mass of first block = M1 = 1.150 kg

Velocity of 2nd block after collision =V2 = 0.602 m/s

Mass of 2nd block = M2 = 1.529 kg

a) From conservation of moentum , Momentum of bullet before collision = oentum of bullet + momentum of first block after collision

(0.0048)(357) = (1.150)(0.602) + (0.0048)vbf

0.0048 vbf = 0.709 => v_bf = 0.709 / 0.0048 = 147.8 m/s =bullet's velocity after collision

Bullet embeds in the 2nd block and so the 2nd block & bullet with a common velocity.

mvbf =( M2 + m ) V

=> V = (1.529 + 0.0048) / (0.0048 * 147.8) = 2.16 m/s

b) Kinetic energy Before collision = 1/2 mb v^2 = 1/2 (0.0048)(357)^2 = 305.9 J

Kinetic energy After the collision = 1/2 (1.529 + 0.0048)(2.16)^2 = 3.578 J

Kinetic after collision / kinetic energy before collision = 3.578 / 305.9 = 0.0117

Final kinetic energy = 1.17% of the initial kinetic energy