Kinetic Energy, Force, and Collisions

[Soaring Crane]

A 49 g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.5 ms. The ball rebounds elastically and returns to its original height. The time interval for a round trip is 7.00 s. In this situation, the average force exerted on the ball during contact with the plate is closest to:
A.4490 N----b. 3360 N------------c. 7850 N-----d. 6720 N------e. 5590 N

If t_total = 7.00 s and time of collision is 0.5 ms*(1 s/1000 ms) = 5*10^-4 s, then 7.00 s – 5*10^-4 s = 6.9995 s, which is the time that is not dedicated to the contact time.

Since the ball rebounds, it takes 6.9995 s/2 = 3.49975 s for the ball to travel from rest to the plate another 3.49975 s for the ball to travel from the plate back to the original height.

F = m*(delta v)/t_contact

To find the ball’s velocity, I though of the conservation of energy,
Where v = sqrt(2*g*h).

I also thought of the kinematic equation v_f = v_o + a*t in which a = 9.8 m/s^2 if the ball is dropped.

Now, would v = 9.8 m/s^2*(3.49975 s) = 34.29755 m/s? (same value if h = 0.5a*t^2 used)

F = (.049 kg)*(34.29755 m/s)/(5*10^-4 s) = 3361.2 J ?




A 18 g bullet is shot vertically into an 10 kg block. The block lifts upward 9 mm. The bullet penetrates the block in a time interval of 0.001 s. Assume the force on the bullet is constant during penetration. The initial kinetic energy of the bullet is closest to:

a.490 J
b.0.88 J
c.250 J
d.330 J
e.0.0016 J

First off, I was puzzled at why a time interval was given. “Does it possibly have something to do with the impulse?” I thought.

Here is what I ended up doing. Please check to see if the setup is correct.

Inelastic collision: m_bullet*v = (m_bullet + m_block)*v’

For block: For conservation of energy, mgh = 0.5*m*v^2 ??
v = sqrt(2*g*h) = sqrt(2*9.8*0.009 m) = 0.42 m/s

So v_block = v’?

V_bullet = [(m_bullet + m_block)*v’]/[ m_bullet] = [(0.018 kg + 10 kg)*0.42 m/s]/[0.018 kg] = 233.753 m/s

KE = (mv^2)/2 = 0.5*(0.018 kg)*(233.753 m/s)^2 = 492 J (choice a.)

Thanks.

 

[PhanthomJay]

Yes, nice work on both. In the first problem, i wouldn't worry about that 0.5ms in comparison to the 7000ms flight time . In question 2, the impact time is not needed, but what really is not needed is the assumption that the force on the bullet is constant, since it most definitely is not.

 

[kyle8921]

Your calculations seem to be correct. The time interval is given because it is important to consider the duration of the collision in order to calculate the average force exerted on the bullet. The impulse-momentum theorem states that the change in momentum of an object is equal to the force applied multiplied by the time interval over which the force acts. In this case, the bullet experiences a constant force for 0.001 s during penetration, so the average force exerted on the bullet can be calculated using the formula F = (m_bullet * v) / t.

Using the values you calculated, the average force exerted on the bullet during penetration is approximately 492 N. This is the force that causes the bullet to penetrate the block and lift it upward. The initial kinetic energy of the bullet can also be calculated using the formula KE = (m_bullet * v^2) / 2, which gives a value of 492 J, as you correctly calculated. Therefore, the answer is a. 490 J.

It's great to see that you used multiple equations and concepts to arrive at your answer. Keep up the good work!