Question

A 5.87-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 1238 g, and its velocity is +0.631 m/s after the bullet passes through it. The mass of the second block is 1623 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

let

m = 5.87 g = 0.00587 kg
u = 348 m/s
m1 = 1238 g = 1.238 kg
v1 = 0.631 m/s (after the collision)
m2 = 1623 g = 1.623 kg

let u1 is the speed of bullet after passing through the first block.

a) Apply conservation of momentum when bullet makes collision with first block.

m*u = m*u1 + m1*v1

==> u1 = (m*u - m1*v1)/m

= (0.00587*348 - 1.238*0.631)/0.00587

= 215 m/s

let v2 is the speed of the second block when bullet embeded with it.

now apply conservation of momentum

m*u1 = (m + m2)*v2

==> v2 = m*u1/(m + m2)

= 0.00587*215/(0.00587 + 1.623)

= 0.775 m/s <<<<<<<<<<----------------------------Answer

b) KEi = (1/2)*m*u^2

= (1/2)*0.00587*348^2

= 355.4 J

KEf = (1/2)*m1*v1^2 + (1/2)*(m + m2)*v2^2

= (1/2)*1.238*0.631^2 + (1/2)*(0.00587 + 1.623)*0.775^2

= 0.7356 J

KEf/KEi = 0.7356/355.4

= 0.00207 <<<<<<<<<<<<---------------------Answer