Question

A 2.5kg wooden block is held by a support on the surface of a frictionless incline. A 20g bullet is shot at the block from below, in a direction parallel to the incline, at a speed of 550m/s. (a) Calculate the velocity of the center of mass of the two bodies before the collision. (b) If the bullet lodges into the block, find the block’s velocity immediately after the collision. (c) At what maximum height (measured from the initial position) will the block climb? (d) If the incline makes an angle of 25° with the horizontal, how far will the block climb before reversing direction?

Answer 1

here,

mass of bullet , m = 20 g = 0.02 kg

mass of block , M = 2.5 kg

the initial speed of bullet , u = 550 m/s

a)

the velocity of the center of mass of the two bodies before the collision , vcm = m * u /( m + M)

vcm = 0.02 * 550 /( 0.02 + 2.5) = 4.37 m/s

b)

let the final speed of block after the collison be v

using conservation of momentum

m * u = (m + M) * v

0.02 * 550 = ( 0.02 + 2.5) * v

solving for v

v = 4.37 m/s

c)

the maximum height , hm = v^2 /( 2*g)

hm = 4.37^2 /( 2 * 9.81 )= 0.97 m

d)

theta = 25 degree

the distance travelled after the collison , s = hm/sin(theta)

s = 0.97 /sin(25) = 2.3 m