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A 65.0 - kg person throws a 0.045 0 - 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 arethe velocities of the two people after the snowball is exchanged? Disregard friction between the skates and the ice.

$v _ { T h r o u s r } = 2.48 \mathrm { m } / \mathrm { s }$$v _ { \text {Calcher } } = 0.0225 \mathrm { m } / \mathrm { s }$

04:14

Salamat A.

Physics 101 Mechanics

Chapter 6

Momentum, Impulse, and Collisions

Energy Conservation

Moment, Impulse, and Collisions

University of Michigan - Ann Arbor

University of Washington

Hope College

McMaster University

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in this problem on the topic of momentum, impulse and collisions. A 65 Kg person is throwing a snowball of mass, 0.045 Kg forward With a speed little to the ground of 13 m/s. A second person with a massive 60 kg then catches the snowball. And if we are told that both people are on skates and we are told to disregard the friction between these gates and the ice. And we're also told that the first person is initially moving forward to the speed of 2.5 m/s while the second is addressed. We want to calculate the velocities of the two people after they exchanged the snowball, so we'll consider it through our first with the velocity after the thrower in VT, we can apply the conservation of momentum as follows. So the conservation of momentum for the thrower is 65 kg. And the initial and the velocity V. T plus 0.0 for five Kg, which is the mass of the snowball, times its velocity that he meters the second must equal to 65 Plus 0.045 kg times the initial speed when the Before they throw it and throw the snowball, which is 2.5 m/s, which means we can rearrange and find the speed of the thrower after they throw the snowball, VT B 2.48 meters per second. So that's the final speed of the throw of the snowball. We can then apply the conservation of momentum to the catcher we see and we get 60 kg. The mess of the catch up plus The mass of the snowball, 0.0 for five Kg times the speed of the catcher. VC is equal to 0.0 four or five kg times 30 m/s, which is the initial momentum of the snowball before the catcher catches it Plus zero, since the catcher is initially addressed, meaning that the speed of the catcher after catching the snowball, which is V. C. Is to 0.25 Times 10 to the -2 m A Second.

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