A 5.75 g bullet is fired with a velocity of 1.50 x 102 m/s toward a stationary solid block resting on a frictionless surface. The bullet embeds but the block does not move. 1. What is the change in momentum of the bullet if it embeds in the block? 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?

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

it should be -8.6 cm hope this out

kepler used the data from tycho blahe (his mentor) to construct the laws of planetary motion.