Question

I have three questions. John has to hit a bottle with a ball to win a prize. He throws a 0.4 kg ball with a velocity of 18 m/s. It hits a 0.2 kg bottle, and the bottle flies forward at 25 m/s. How fast is the ball traveling after hitting the bottle?
2. Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of woof is sitting on a frictionless surface, so it moves freely when the bullet hits it). The wood block is initially at rest. The bullet is traveling 300 m/s when it hits the wood block and sticks inside it. Now the bullet and the woof block move together as one object. How fast are they traveling?

3. A big league hitter attacks a fastball. The ball has a mass of 0.16 kg. It is pitched at 38 m/s. After the player hits the ball, it is now traveling 44 m/s in the opposite direction. The impact lasted 0.002 seconds. How big of a force did the ballplayer put on that ball?

Answer 1

1. 5.5 m/s

We can solve the problem by applying the law of conservation of momentum. The total momentum before the collision must be equal to the total momentum after the collision, so we have:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where

m1 = 0.4 kg is the mass of the ball

u1 = 18 m/s is the initial velocity of the ball

m2 = 0.2 kg is the mass of the bottle

u2 = 0 is the initial velocity of the bottle (which is initially at rest)

v1 = ? is the final velocity of the ball

v2 = 25 m/s is the final velocity of the bottle

Substituting and re-arranging the equation, we can find the final velocity of the ball:

$v_1 = \frac{m_1 u_1 - m_2 v_2}{m_1}=\frac{(0.4 kg)(18m/s)-(0.2 kg)(25 m/s)}{0.4 kg}=5.5 m/s$

2. 22.2 m/s

We can solve the problem again by using the law of conservation of momentum; the only difference in this case is that the bullet and the block, after the collision, travel together at the same speed v. So we can write:

$m_1 u_1 + m_2 u_2 = (m_1 +m_2) v$

where

m1 = 0.04 kg is the mass of the bullet

u1 = 300 m/s is the initial velocity of the bullet

m2 = 0.5 kg is the mass of the block

u2 = 0 is the initial velocity of the block (which is initially at rest)

v = ? is the final velocity of the bullet+block, which stick and travel together

Substituting and re-arranging the equation, we can find the final velocity of bullet+block:

$\frac{m_1 u_1}{m_1 +m_2}=\frac{(0.04 kg)(300 m/s)}{0.04 kg+0.5 kg}=22.2 m/s$

3. 6560 N

The impulse exerted on the ball is equal to its change in momentum:

$I=\Delta p$ (1)

The impulse can be rewritten as product between force and time of collision:

$I=F \Delta t$

while the change in momentum of the ball is equal to the product between its mass and the change in velocity:

$\Delta p = m\Delta v = m(v_f -v_i)$

So, eq.(1) becomes

$F \Delta t = m(v_f -v_i)$

where:

F = ? is the unknown force

$\Delta t = 0.002 s$ is the duration of the impact

m = 0.16 kg is the mass of the ball

$v_f = 44 m/s$ is the final velocity of the ball

$v_i = -38 m/s$ is its initial velocity (we must add a negative sign, since it is in opposite direction to the final velocity)

So, by using the equation, we can find the force:

$F=\frac{m (v_f -v_i)}{\Delta t}=\frac{(0.16 kg)(44 m/s-(-38 m/s))}{0.002 s}=6560 N$