A 2 kg cart rolls across the floor and runs into the wall and sticks. Before the collision the cart has a velocity of 4 m/s in the x direction. The collision took place in 0.1 s.

1) What was the cart's momentum before the collision?
2) What was the cart's KE before the collision?
3) What was the momentum after the collision?
4)What was the change in the cart's momentum?
5) What was the impulse delivered to the car by the wall?
6) What was the average force exerted on the wall by the car?

PLEASE HELP!!!
its a direct answer.

1) The cart's momentum before the collision can be calculated using the equation p = m * v, where p is the momentum, m is the mass, and v is the velocity. Since the mass of the cart is 2 kg and the velocity is 4 m/s, the momentum before the collision is:

p = 2 kg * 4 m/s = 8 kg*m/s

2) The cart's kinetic energy (KE) before the collision can be calculated using the equation KE = 0.5 * m * v^2, where m is the mass and v is the velocity. Plugging in the values, we get:

KE = 0.5 * 2 kg * (4 m/s)^2 = 16 J

3) The momentum after the collision can be determined by considering that the cart sticks to the wall and therefore comes to rest. Therefore, the momentum after the collision is zero.

4) The change in momentum can be calculated by subtracting the momentum after the collision from the momentum before the collision:

Change in momentum = Momentum before - Momentum after
= 8 kg*m/s - 0 kg*m/s
= 8 kg*m/s

5) The impulse delivered to the cart by the wall can be calculated using the equation Impulse = Change in momentum. Therefore, the impulse is:

Impulse = 8 kg*m/s

6) The average force exerted on the wall by the cart can be calculated using the equation Impulse = Force * Time. Given that the collision duration is 0.1 s, we can rearrange the equation to solve for the average force:

Force = Impulse / Time
= 8 kg*m/s / 0.1 s
= 80 N

Therefore, the average force exerted on the wall by the cart is 80 N.

Sure, I'd be happy to help you answer these questions step by step! Let's start with the first question:

1) What was the cart's momentum before the collision?

Momentum (p) is calculated by multiplying an object's mass (m) by its velocity (v). In this case, the mass of the cart is given as 2 kg, and its velocity is given as 4 m/s in the x direction. So, to find its momentum before the collision, we can use the following formula:

Momentum (before collision) = mass * velocity

Plugging in the values, we get:

Momentum (before collision) = 2 kg * 4 m/s
= 8 kg m/s

Therefore, the cart's momentum before the collision is 8 kg m/s.

2) What was the cart's KE before the collision?

Kinetic energy (KE) is given by the formula KE = (1/2) * mass * velocity^2. In this case, we are given the mass of the cart as 2 kg and its velocity as 4 m/s. Plugging in the values, we can calculate the kinetic energy before the collision as follows:

KE (before collision) = (1/2) * mass * velocity^2
= (1/2) * 2 kg * (4 m/s)^2
= 16 J

Hence, the cart's kinetic energy before the collision is 16 Joules.

3) What was the momentum after the collision?

Since the cart sticks to the wall after the collision, its velocity becomes zero. Momentum (p) is still calculated by multiplying an object's mass (m) by its velocity (v). Since velocity is zero after the collision, the momentum after the collision will also be zero.

Therefore, the momentum after the collision is 0 kg m/s.

4) What was the change in the cart's momentum?

The change in momentum (Δp) is simply the difference between the initial momentum and the final momentum. In this case, the initial momentum before the collision was 8 kg m/s (as calculated in question 1) and the final momentum after the collision is 0 kg m/s (as calculated in question 3).

Therefore, the change in the cart's momentum is:

Change in momentum (Δp) = Final momentum - Initial momentum
= 0 kg m/s - 8 kg m/s
= -8 kg m/s

The negative sign indicates a decrease in momentum.

5) What was the impulse delivered to the cart by the wall?

Impulse (J) is the change in momentum and can be calculated using the formula J = Δp = F * Δt, where Δp is the change in momentum, F is the force exerted, and Δt is the time interval.

In this case, we are given the time interval Δt = 0.1 s. We already calculated the change in momentum (Δp = -8 kg m/s) in question 4.

So, rearranging the equation, we get:

Impulse (J) = Δp = F * Δt
= (-8 kg m/s) = F * 0.1 s

Solving for F, we find:

F = (-8 kg m/s) / 0.1 s
= -80 N

The negative sign indicates that the impulse was delivered in the opposite direction of the initial momentum.

Therefore, the impulse delivered to the cart by the wall is -80 Newton-seconds (N·s).

6) What was the average force exerted on the wall by the cart?

The average force (F) exerted on the wall can be calculated using the formula:

F = Impulse / time

In this case, we found the impulse to be -80 N·s, and we are given the time interval Δt = 0.1 s.

Plugging in the values, we get:

Force (F) = (-80 N·s) / 0.1 s
= -800 N

Therefore, the average force exerted on the wall by the cart is -800 Newtons (N).

Please note that the negative sign indicates that the force is directed opposite to the initial motion of the cart.

I hope that helps! Let me know if you have any further questions.