Change in Momentum and Impulse AP Physics 1 FRQ – Exam Style Questions etc.
Change in Momentum and Impulse AP Physics 1 FRQ
Unit 4: Linear Momentum
Weightage : 10-15%
Exam Style Practice Questions, Change in Momentum and Impulse AP Physics 1 FRQ
Question
Two objects of masses Ml = 1 kilogram and M2 = 4 kilograms are free to slide on a horizontal frictionless surface. The objects collide and the magnitudes and directions of the velocities of the two objects before and after the collision are shown on the diagram above. (sin 37° = 0.6, cos 37° = 0.8, tan 37º = 0.75)
a. Calculate the x and y components (px and py, respectively) of the momenta of the two objects before and after the collision, and write your results in the proper places in the following table.
b. Show, using the data that you listed in the table, that linear momentum is conserved in this collision.
c. Calculate the kinetic energy of the two-object system before and after the collision.
d. Is kinetic energy conserved in the collision?
▶️Answer/Explanation
Ans:
a) Before the collision there is only an x direction momentum of mass M1 … px = m1v1x = 16, all the rest are 0
After the collision, M1 has y direction momentum = m1v1fy = 12 and M2 has x and y direction momentums.
Using trig to find the x and y velocities of mass M2 … vx = 5 cos 37 = 3, and vy = 5 sin 37 = 3.75.
Then plug into mv to get each x and y momentum after.
| M1 = 1 kg | M2 = 4 kg | |||
| px (kg m /s) | py (kg m /s) | px (kg m /s) | py (kg m /s) | |
| Before | 16 | 0 | 0 | 0 |
| After | 0 | -12 | 16 | 12 |
b) SUM = 16 -12 16 12
When adding x’s before they = x’s after 16=16, when adding y’s before they equal y’s after |-12|=12
c) Kinetic Energy Before Kinetic Energy After
K = ½ m1v1ix2 K = ½ m1v1fy2 + ½ m2v22
K = ½ (1)(16)2 = 128 J K = ½ (1)(12)2 + ½ (4)(5)2 = 122 J
d) From above, K is not conserved.
Question
A 0.20 kg object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown in the graph above. The object starts from rest at displacement x = 0 and time t = 0 and is displaced a distance of 20 m. Determine each of the following.
a. The acceleration of the particle when its displacement x is 6 m.
b. The time taken for the object to be displaced the first 12 m.
c. The amount of work done by the net force in displacing the object the first 12 m.
d. The speed of the object at displacement x = 12 m.
e. The final speed of the object at displacement x = 20 m.
f. The change in the momentum of the object as it is displaced from x = 12 m to x = 20 m
▶️Answer/Explanation
Ans:
a) The force is constant, so simple Fnet = ma is sufficient. (4) = (0.2) a a = 20 m/s2
b) Use d = vit + ½ a t2 12 = (0) + ½ (20) t2 t = 1.1 sec
c) W = Fd W = (4 N) (12 m) = 48 J
d) Using work energy theorem W = ∆K (Ki = 0) W = Kf – Ki
W = ½ m vf2
Alternatively, use vf 2 = vi2 + 2 a d 48J = ½ (0.2) (vf2) vf = 21.9 m/s
e) The area under the triangle will give the extra work for the last 8 m
½ (8)(4) = 16J + work for first 12 m (48J) = total work done over 20 m = 64 J
Again using work energy theorem W = ½ m vf2 64 J = ½ (0.2) vf2 vf = 25.3 m/s
Note: if using F = ma and kinematics equations, the acceleration in the last 8 m would need to be found using the average force over that interval.
f) The momentum change can simply be found with ∆p = m∆v = m(vf – vi) = 0.2 (25.3 – 21.9) = 0.68 kg m/s