Conservation of Linear Momentum AP Physics 1 MCQ – Exam Style Questions etc.
Conservation of Linear Momentum AP Physics 1 MCQ
Unit 4: Linear Momentum
Weightage : 10-15%
Exam Style Practice Questions, Conservation of Linear Momentum AP Physics 1 MCQ
Question
An object of mass \(m_{1}\) experiences a linear, elastic collision with a stationary object of unknown mass. In addition to \(m_{1}\), what is the minimum necessary information that would allow you to determine the mass of the second object?
(A) The final speed of object 1
(B) The initial speed of object 1
(C) The final speed of object 2
(D) Any 2 of the above values
Answer/Explanation
Ans:D
For an elastic collision, both kinetic energy and momentum will be conserved. Writing these statements as equations gives
\(\frac{1}{2}m_{1}v^{2}_{1,0}+\frac{1}{2}m_{2}v^{2}_{2,0}=\frac{1}{2}m_{1}v^{2}_{1,f}+\frac{1}{2}m_{2}v^{2}_{2,f}\)
\(m_{1}v_{1,0}+m_{2}v_{2,0}=m_{1}v_{1,f}+m_{2}v_{2,f}\)
You’re told \(m_{1}\) is a given in the question, and \(v_{2,0}\) is also known to be 0 m/s since the object started at rest. This leaves \(m_{2}\), \(v_{1,0}\), \(v_{1,f}\), and \(v_{2,f}\) as unknowns. With two equations, you can have two unknowns. Therefore, (D) is correct.
Question
Two pucks are firmly attached by a stretched spring and are initially held at rest on a frictionless surface, as shown above. The pucks are then released simultaneously. If puck I has three times the mass of puck II, which of the following quantities is the same for both pucks as the spring pulls the two pucks toward each other?
(A) Speed (B) Magnitude of acceleration (C) Kinetic energy (D) Magnitude of momentum
Answer/Explanation
Ans:D
Solution: Since the momentum before is zero, the momentum after must also be zero. Each mass must have equal and opposite momentum to maintain zero total momentum.
Question
The two blocks of masses M and 2M shown above initially travel at the same speed v but in opposite directions. They collide and stick together. How much mechanical energy is lost to other forms of energy during the collision?
(A) 1/2 M v2 (B) 3/4 M v2 (C) 4/3 M v2 (D) 3/2 M v2
Answer/Explanation
Ans:C
Solution: Perfect inelastic collision. m1v1i + m2v2i = mtot(vf) … Mv + (– 2Mv) = (3M) vf gives vf = v/3. Then to find the energy loss subtract the total energy before – the total energy after
[ ½ Mv2 + ½ (2M)v2 ] – ½ (3M) (v/3)2 = 3/6 Mv2 + 6/6 Mv2 – 1/6 Mv2
Question
As shown in the top view, a disc of mass m is moving horizontally to the right with speed v on a table with negligible friction when it collides with a second disc of mass 2m. The second disc is moving horizontally to the right with speed v/2 at the moment before impact. The two discs stick together upon impact. The kinetic energy of the composite body immediately after the collision is
(A) (1/6)mv2 (B) (1/2)mv2 (C) 2/3mv2 (D) (9/8)mv2
Answer/Explanation
Ans:C
Solution: Perfect inelastic collision. m1v1i + m2v2i = mtot(vf) … (m)(v) + (2m)(v / 2) = (3m)vf
Question
A running back (m = 85 kg) running at 1.5 m/s is tackled from the side by another player (m = 75 kg) running perpendicularly to the running back’s original heading at (1.75 m/s). What is the resulting speed of the two entangled players just after the
tackle?
(A) 2.2 m/s
(B) 0.25 m/s
(C) 1.6 m/s
(D) 1.1 m/s
Answer/Explanation
Ans:(D)
Total momentum must be conserved through the collision. So we need to find only the magnitude of the initial momentum of the system. Each player is originally
running perpendicularly to the other. So their individual momentums are different components. Imagine one as an x-component and the other as a y-component. Therefore,
the resulting magnitude is found using the Pythagorean theorem. Dividing by total mass yields the speed of the combined players after impact:
