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AP Momentum HW 1 2. Starting from rest, Object 1 is subject to a 12 N force for 2.0 s. Object 2, with twice the mass, is subject to a 15 N force for 3.0 s. Which object has the greater final speed? Explain. 24
45 22.5
while v f 1 =
, so final velocity for the first J1 = 24Ns and J 2 = 45Ns Both start at rest, so v f 1 =
=
m
2m
m
object will be greater. 4. Two pucks, of mass m and 4m lie on a frictionless table. Equal forces are used to push both pucks forward a distance of 1 m. a. Which puck takes longer to travel the distance? Explain. F
F
Remember Newton’s II law: a1 = while a2 =
, so a1 = 4a2 . If they both start from rest and the m
4m
1
2
acceleration is constant, we can find the times using Δx = a ( Δt ) . With the 4x the acceleration, the first 2
object is take half the time of the second object, since they both travel the same distance: 2 ⋅1
2 ⋅1
Δt =
for the second object and Δt =
for the second object (half the time, once you square root, a
4a
you can through in numbers if you like to check). b. Which puck has the greater momentum upon completing the distance? Explain. For both, Δp = FΔt, since the force is the same, a greater time for the heavier object will result in a greater change of momentum. Problem 3 A student throws a 120 g snowball at 7.5 m/s at the side of the schoolhouse, where it hits and sticks. What is the magnitude of the average force on the wall if the duration of the collision is 0. 15s? Use impulse-­‐momentum theorem: FΔt = m(v f − vi ) . The snowball will naturally stop when it hits the wall, so: F =
.120 ⋅(0 − 7.5)
15
Problem 7 A 2.0 kg object is moving to the right with a speed of 1.0 m/s when it experiences the force shown in Figure P7A. a. What are the object’s speed and direction after the force ends? b. Answer the question for the force shown in Figure P7b. It’s not quite clear in the diagram, but suppose the time is 1/2s, then the first impulse is 1 N⋅s and the second is -­‐1N⋅s (negative). Initial momentum of the object is 2 kg⋅m/s. And the change is positive 1 kg⋅m/s (equals to impulse), making its final momentum 3 kg⋅m/s and final velocity to the right at 1.5 m/s. With the second force, impulse and change in momentum is -­‐
1kg⋅m/s, making final momentum 1 kg⋅m/s and velocity 0.5 m/s, still to the right. Problem 8 A 60 g tennis ball with an initial speed of 32 m/s hits a wall and rebounds with the same speed. Figure P8 shows the force of the wall on the ball during the collision. What is the value of Fmax, the maximum value of the contact force during the collision? Impulse-­‐momentum theorem: FΔt = m(v f − vi ) , and the area under the graph (the area of the trapezoid) equals impulse. So, if you find Δp, you’ll know FΔt. Area of the trapezoid is half of the sum of the two bases, times the height. The height is Fmax. Once you find Δp, use the formula for the are of a trapezoid to find Fmax (the answer for Δp should be 3.84 and Fmax = 0.96 N)