(Serway College Physics 9th Edition Problem 27. Will not find the textbook even

Question

(Serway College Physics 9th Edition Problem 27. Will not find the textbook even though its there.)

A 65.0-kg person throws a 0.0450-kg snowball forward with a ground speed of 30.0 m/s. A second person, with a mass of 60.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.50 m/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.

Explanation / Answer

This is a momentum question. first person - - - - - - - - m1*v1 = m*v + m1*v2 m1 is the mass of the person. m is the mass of the snowball. v is the speed of the snowball. v2 is the speed of the person after throwing the snowball. m1 = 65kg v1 = 2.5 m/s m = 0.0450 kg v = 30 m/s v2 = ??? We have to assume that snowball and the person are going in the same direction. 65*2.5 = 0.0450*30 + 65*v2 v2 = 2.47 m/s Person 2 - - - - - - - The initial speed of the snowball and its mass is still 0.0450 and 30 m/s The second person catches the snowball and is pushed back by the impact. m*v = M3*v3 0.0450 * 30 = 60 v3 0.0225 m/s = v3. Problem 2 ======== Whose body will experience the broken bone? The first skater is going to slow from 10m/s to 5 m/s. This will happen in 0.1 s F = m*(vf - vi)/t F = 75*(5 - 10)/0.1 F = - 75*5/0.1 F = - 3750 N now a bone won't snap. The second scater will experience the same thing. F = 75(5 - 0)/0.1 F = 3750 No emergency here either. Notice the sign difference. One is giving away the Force and one taking it on.

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