How Do You Calculate Momentum and Kinetic Energy of a Moving Truck?

[ChessFanatic]

Homework Statement

A 2000 kg truck is moving at a speed of 30 m/s. It hits a haystack and stops in 80m.
1a. What is the original momentum of the truck?
1b. What is the final momentum of the truck?
2a. What is the original kinetic energy of the truck?
2b. What is the final kinetic energy of the truck?

I'm not sure how to do these, and I've started some, but stuck on the last two. Also, am I correct in assuming final velocity to be 0 m/s?

Homework Equations

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f
(1/2)m₁v_1i² + (1/2)m₂v_2i2 = (1/2)m₁v_1f² + (1/2)m₂v_2f²
P=mv

The Attempt at a Solution

1a. P=mv
P=(2000 kg)(30m/s)= 60,000 kg x m/s

1b. P=mv
P=(2000kg)(0 m/s) = 0 kg x m/s

 

[SammyS]

Homework Statement

A 2000 kg truck is moving at a speed of 30 m/s. It hits a haystack and stops in 80m.
1a. What is the original momentum of the truck?
1b. What is the final momentum of the truck?
2a. What is the original kinetic energy of the truck?
2b. What is the final kinetic energy of the truck?

I'm not sure how to do these, and I've started some, but stuck on the last two. Also, am I correct in assuming final velocity to be 0 m/s?

Homework Equations

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f
(1/2)m₁v_1i² + (1/2)m₂v_2i2 = (1/2)m₁v_1f² + (1/2)m₂v_2f²
P=mv

The Attempt at a Solution

1a. P=mv
P=(2000 kg)(30m/s)= 60,000 kg x m/s

1b. P=mv
P=(2000kg)(0 m/s) = 0 kg x m/s

You have done 1a & 1b correctly.

Do you know how to calculate kinetic energy?

 

[ChessFanatic]

The initial kinetic energy of the truck is 900,000 kg x m/s? that seems wrong.. but the final should be 0, since final velocity would be 0 m/s.

 

[haruspex]

The initial kinetic energy of the truck is 900,000 kg x m/s? that seems wrong.. but the final should be 0, since final velocity would be 0 m/s.

kg x m/s is wrong certainly. The number is right if you meant kg m²/s² (better known as Joules). Why did it seem wrong?

 

[Nickie]

2a. KE=(1/2)mv2
KE=(1/2)(2000 kg)(30m/s)2 = 900,000 joules

2b. KE=(1/2)mv2
KE=(1/2)(2000 kg)(0 m/s)2 = 0 joules

You are correct in assuming that the final velocity of the truck is 0 m/s after hitting the haystack and coming to a stop. This is because the problem states that the truck "stops" in 80m, implying that it comes to a complete stop.

Your solutions for 1a and 1b are correct. For 2a and 2b, you have correctly used the equation for kinetic energy, but you need to use the final velocity (0 m/s) instead of the initial velocity (30 m/s). Therefore, the correct solutions for 2a and 2b are:

2a. KE=(1/2)mv2
KE=(1/2)(2000 kg)(30 m/s)2 = 900,000 joules

2b. KE=(1/2)mv2
KE=(1/2)(2000 kg)(0 m/s)2 = 0 joules

You could also use the equation for work (W=F*d) to calculate the kinetic energy in this scenario. Since the truck comes to a complete stop, the work done by the friction force is equal to the initial kinetic energy of the truck. Therefore, the solutions for 2a and 2b can also be written as:

2a. KE = W = F*d = (2000 kg)(30 m/s)(80 m) = 480,000 joules

2b. KE = W = F*d = (2000 kg)(0 m/s)(80 m) = 0 joules

Overall, your solutions are correct and show a good understanding of the concepts of momentum and kinetic energy. Keep up the good work!