Kinetic Energy, Force, and Collisions

In summary, a 49 g steel ball is released from rest and falls vertically onto a steel plate. The ball rebounds elastically and returns to its original height after a round trip time of 7.00 s. The average force exerted on the ball during contact with the plate is closest to 3360 N. In the second conversation, an 18 g bullet is shot vertically into a 10 kg block, lifting it 9 mm with a penetration time of 0.001 s. The initial kinetic energy of the bullet is closest to 490 J.
  • #1
Soaring Crane
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0
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.
 
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  • #2
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.
 
  • #3



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!
 

Related to Kinetic Energy, Force, and Collisions

1. What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is determined by the mass and velocity of the object and is measured in joules (J).

2. How is kinetic energy calculated?

The formula for calculating kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object in kilograms and v is the velocity of the object in meters per second.

3. What is the relationship between kinetic energy and velocity?

Kinetic energy and velocity have a direct relationship. As the velocity of an object increases, its kinetic energy also increases. This means that an object with a higher velocity will have a greater kinetic energy than an object with a lower velocity, assuming they have the same mass.

4. What is the role of force in changing an object's kinetic energy?

Force is necessary to change an object's kinetic energy. When an external force is applied to an object, it can either increase or decrease the object's velocity, thus changing its kinetic energy. This change in kinetic energy is determined by the amount of force applied and the direction in which it is applied.

5. How does a collision affect an object's kinetic energy?

During a collision, there is a transfer of kinetic energy between the objects involved. In an elastic collision, the total kinetic energy of the system remains the same before and after the collision. In an inelastic collision, some of the kinetic energy is lost as heat or sound. In both cases, the kinetic energy of the individual objects may change depending on their masses and velocities.

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