Question

A 6.13-g bullet is moving horizontally with a velocity of 361 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 1233 g, and its velocity is 0.741 m/s after the bullet passes through it. The mass of the second block is 1646 g. (a) What is the velocity of the second block after the bullet imbeds itself

Answer 1

Answer:

v₃ = 0.786 m/s

Explanation:

Here, we will use the law of conservation of momentum, which states the following:

Total Momentum of System Before Collision =

Total Momentum of System After Collision

m₁u₁ + m₂u₂ + m₃u₃ = m₁v₁ + m₂v₂ + m₃v₃

where,

m₁ = mass of bullet = 6.13 g = 0.00613 kg

m₂ = mass of 1st block = 1233 g = 1.233 kg

m₃ = mass of 2nd block = 1646 g = 1.646 kg

u₁ = speed of first bullet before collision = 361 m/s

u₂ = speed of first block before collision = 0 m/s

u₃ = speed of 2nd block before collision = 0 m/s

v₁ = speed of bullet after collision

v₂ = speed of 1st block after collision = 0.741 m/s

v₃ = speed of 2nd block after collision = ?

Therefore,

(0.00613 kg)(361 m/s) + (1.233 kg)(0 m/s) + (1.646 kg)(0 m/s) = (0.00613 kg)(v₁) + (1.233 kg)(0.741 m/s) + (1.646 kg)(v₃)

2.2129 kg m/s + 0 kg m/s + 0 kg m/s - 0.9136 kg m/s = (0.00613 kg)(v₁) + (1.233 kg)(0.741 m/s) + (1.646 kg)(v₃)

1.2992 kg m/s = (0.00613 kg)(v₁) + (1.646 kg)(v₃)

since, the bullet is embedded in 2nd block after collision. Thus, there velocities will become same. (v₁ = v₃)

Therefore,

1.2992 kg m/s = (0.00613 kg)(v₃) + (1.646 kg)(v₃)

v₃ = (1.2992 kg m/s)/(1.6521 kg)

v₃ = 0.786 m/s