Two pucks moving on a frictionless air table are about to collide, as shown. The 1.5 kg puck is moving directly east at 2.0 m/s.The 4.0 kg puck is moving directly north at 1.0 m/s.

Answer: B

Simply add the energies ½ (1.5)(2)^{2} + ½ (4)(1)^{2}

Answer: C

Total momentum before must equal total momentum after. Before, there is an x momentum of (2)(1.5)=3 and a y momentum of (4)(1)=4 giving a total resultant momentum before using

the Pythagorean theorem of 5. The total after must also be 5.

A disk slides to the right on a horizontal, frictionless air table and collides with another disk that was initially stationary. The figures below show a top view of the initial path I of the sliding disk and a hypothetical path H for each disk after the collision. Which figure shows an impossible situation?

Answer: B

This is a 2D collision. Before the collision, there is no y momentum, so in the after condition the

y momenta of each disk must cancel out. In choice B, both particles would have Ymomentum downwards making a net Y momentum after which is impossible

Two objects of mass 0.2 kg and 0.1 kg, respectively, move parallel to the x-axis, as shown on the left. The 0.2 kg object overtakes and collides with the 0.1 kg object. Immediately after the collision, the y-component of the velocity of the 0.2 kg object is 1 m/s upward. What is the y-component of the velocity of the 0.1 kg object immediately after the collision'?

2 m/s downward (B) 0.5 m/s downward (C) 0 m/s (D) 0.5 m/s upward (E) 2 m/s upward

2D collision. Analyze the y direction. Before py = 0 so after py must equal 0.

0 = m1v1fy + m2v2fy 0 = (0.2)(1) + (0.1)(V2fy)

Two particles of equal mass mo, moving with equal speeds vo. Along paths inclined at 60° to the x-axis as shown, collide and stick together. Their velocity after the collision has magnitude

Answer: B

2D collision. The y momentums are equal and opposite and will cancel out leaving only the x momentums which are also equal and will add together to give a total momentum equal to twice the x component momentum before hand. pbefore = pafter 2movocos60 = (2mo) vf

initially as shown. The magnitude of the momentum change of the ball is

0 (B) mv (C) 2mv (D) 2mv sinθ (E) 2mv cosθ

A 2d collision must be looked at in both x-y directions always. Since the angle is the same and the v is the same, vy is the same both before and after therefore there is no momentum change in the y direction. All of the momentum change comes from the x direction.

vix = v cos θ and vfx = –v cos θ. Δp = mvfx – mvix … – mv cos θ – mv cos θ

2 kg ball collides with the floor at an angle θ and rebounds at the same angle and speed as shown above. Which of the following vectors represents the impulse exerted on the ball by the floor?

Answer: E

Since the angle and speed are the same, the x component velocity has been unchanged which means there could not have been any x direction momentum change. The y direction velocity was reversed so there must have been an upwards y impulse to change and reverse the velocity.