Answer:
1. -0.863 kgm/s 2. -1.438 kgm/s
Explanation:
1. What is the change in momentum of the bullet if it embeds in the block?
Since the block does not move, the velocity of the bullet after hitting the block , v is zero. That is v = 0 m/s
Now, the momentum change of the bullet ΔP = m(v - u) where m = mass of block = 5.75 g = 5.75 × 10⁻³ kg, u = initial velocity of bullet = 1.50 × 10² m/s and v = final velocity of bullet after hitting the block = 0 m/s (since it embeds in the block and the block does not move).
So, ΔP = m(v - u)
= 5.75 × 10⁻³ kg(0 m/s - 1.50 × 10² m/s)
= 5.75 × 10⁻³ kg(- 1.50 × 10² m/s)
= -8.625 × 10⁻¹ kgm/s
= -0.8625 kgm/s
≅ -0.863 kgm/s
2. What is the change in momentum of the bullet if it bounces off the block in the opposite direction with a speed of 100 m/s?
If it bounces off the block in the opposite direction with a speed of 100 m/s, then its final velocity is v = -100 m/s.
So, our momentum change ΔP' = m(v - u) where m = mass of block = 5.75 g = 5.75 × 10⁻³ kg, u = initial velocity of bullet = 1.50 × 10² m/s and v = final velocity of bullet after hitting the block = -100 m/s = -1 × 10² m/s
So, ΔP = m(v - u)
= 5.75 × 10⁻³ kg(-1 × 10² m/s - 1.50 × 10² m/s)
= 5.75 × 10⁻³ kg(-2.50 × 10² m/s)
= -14.375 × 10⁻¹ kgm/s
= -1.4375 kgm/s
≅ -1.438 kgm/s
