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