Question

In Figure (1), a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.42 kg) and embeds itself in block 2 (mass 1.88 kg). The blocks end up with speeds v1 = 0.610 m/s and v2 = 1.31 m/s (see Figure (2)). Neglecting the material removed from block 1 by the bullet, find the speed of the bullet as it (a) leaves and (b) enters block 1.

Answer 1

Mass of the bullet = m = 3.5 g = 3.5 x 10^-3 kg

Mass of the first block = m₁ = 1.42 kg

Mass of the second block = m₂ = 1.88 kg

Initial speed of the bullet = V_a

Speed of the first block after the leaves it = V₁ = 0.61 m/s

Speed of the bullet as it leaves the first block = V_b

Speed of the second block and the bullet after the collision of the bullet and the second block = V₂ = 1.31 m/s

By conservation of linear momentum for the collision of the bullet with the second block,

mV_b = (m + m₂)V₂

(3.5x10^-3)V_b = (3.5x10^-3 + 1.88)(1.31)

V_b = 704.97 m/s

By conservation of linear momentum for the collision of the bullet with the first block,

mV_a = mV_b + m₁V₁

(3.5x10^-3)V_a = (3.5x10^-3)(704.97) + (1.42)(0.61)

V_a = 952.46 m/s

a) Speed of the bullet as it leaves the first block = 704.97 m/s

b) Speed of the bullet as it enter the first block = 952.46 m/s