1 . Find the final velocity of the two balls if the ball with velocity v 2 i = -

Question

1. Find the final velocity of the two balls if the ball with velocity v2i = -20.5 cm/s has a mass equal to half that of the ball with initial velocity v1i = +26.2 cm/s.

2. A billiard ball rolling across a table at 1.25 m/s makes a head-on elastic collision with an identical ball. Find the speed of each ball after the collision when each of the following occurs.

(b) The second ball is moving toward the first at a speed of 1.15 m/s.

**1st ball??: 1.15 m/s The response you submitted has the wrong sign.

2nd ball: 1.25 m/s

3. A 68.0 kg person throws a 0.0500 kg snowball forward with a ground speed of 33.0 m/s. A second person, with a mass of 59.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.20m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard the friction between the skates and the ice.

4.

v1f = ____ cm/s v2f = ____ cm/s Because no external force acts, the collision does not change the total momentum of the system of two particles. We set the total momentum before collision to the total momentum afterward: m1v1 m2V2 m1 m2)vf The final velocity of the two objects joined together is given by the previous momentum conservation equation: m1V1 2V2 (1) vf m1 m2 Using the values of this exercise, we get 18 m/s) (19 m/S m/s 2.59 Vf 18 kg) (19 kg) (B) Determine the change in total mechanical energy. While the forces that the two objects exert on each other cannot change their total momentum, they can change the total kinetic energy in an inelastic collision such as the one being considered. All of the mechanical energy in the problem is kinetic energy. The change in kinetic energy is then AK 1/2 (m1 m2)v 1/2m2v22) At this point, we can either substitute the numerical values to evaluate AK, or we could substitute in Equation (1) for vf into the expression for AK and simplify the expression before doing the calculation. The first procedure is easier if only a numerical answer is needed, while the second is more useful in exploring how the loss in kinetic energy depends on the velocities and masses of the colliding objects. Directly substituting in numerical values gives AK Ve(18 kg 19 Vf m/s)2 [Va(18 kg)(-1 m/s)2 Vi(19 kg)(6 m/s)2] 18.97 X J. Following the second procedure instead leads to m1 m2 AK (v1 v2) 2 m1 m2

Explanation / Answer

V1f = (m1-m2)*V1i / (m1+m2) ] + [(2*m2*V2i)/(m1+m2)]

V1f = [((m-0.5m)*26.2) / (m+0.5m)] - [(2*0.5*m*20.5)/(m+0.5m)]

V1f = 8.73 - 13.66

V1f = -4.93 cm /sec

V2f = [2*m1*V1i/(m1+m2) ] - [(m1-m2)*v2i)/(m1+m2)]

v2f = (2*m*26.2/(m+0.5m)) - ((m-0.5m)*20.5/(m+0.5m))

V2f = 34.9 - 6.83 = 28.07 cm /sec

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