Question

You have recently purchased a gun and want to test it out. Although probably not the best idea, you decide to fire it horizontally into a large block of wood resting on the ground. After the bullet hits the block of wood, it lodges in the block and the two move together. The mass of the bullet is 4.2 grams. The mass of the block is 1.5 kg. After the impact, the block-and-bullet combined object slides 0.24 m along the ground. The coefficient of kinetic friction between the block and the ground is 0.5.

(a) How much energy was lost in the bullet-block combined object due to the friction between the block and the ground?

(b) What was the speed of the bullet-block combined object just after the collision?

(c) What was the speed of the bullet before the collision? (d) How much energy was lost in the collision itself?

Answer 1

here,

mass of bullet , m1 = 4.2 g = 0.0042 kg

mass of block , m2 = 1.5 kg

distance traveled before stopping , s = 0.24 m

the coefficient of friction , uk = 0.5

a)

the energy was lost in the bullet-block combined object due to the friction between the block and the ground, U = uk * (m1 + m2) * g * s

U = 0.5 * (1.5 + 0.0042) * 9.81 * 0.24 J

U = 1.77 J

b)

let the speed of bullet-block combined just after the collison be v

using work energy theorm

initial kinetic energy = U

0.5 * (m1 + m2) * v^2 = 1.77

0.5 * ( 1.5 + 0.0042) * v^2 = 1.77

v = 1.53 m/s

the speed of the bullet-block combined object just after the collision is 1.53 m/s

c)

the speed of the bullet before the collision , u = (m1 + m2) * v /(m1)

u = (1.5 + 0.0042) * 1.53 /(0.0042) m/s

u = 548 m/s

(d)

the energy was lost in the collision itself , KE = KEi - KEf

KE = 0.5 * (m1 * u^2)- 0.5 * (m1 + m2) * v^2

KE = 0.5 * ( 0.0042 * 548^2) - 0.5 * ( 1.5 + 0.0042) * 1.53^2 J

KE = 628.9 J