Question

In: Physics

two skaters collide and embrace, in a completely inelastic collision. boomer, whose mass is 83 kg,...

two skaters collide and embrace, in a completely inelastic collision. boomer, whose mass is 83 kg, is originally moving east with v=6.2km/hr. sooner whose. mass is 55kg, is originally moving north with v=7.8km/hr. (a) what is the velocity of the couple after impact? (b) what is the velocity of the center of mass of two skaters before and aafter collision? (c) what is the fractional change in the Kinect energy of the skaters because of the collision?

Solutions

Expert Solution

In a completely inelastic collision all you need to consider is the conservation of momentum.

Eastward momentum (before collision):
Skater (1) 83 kg* 6.2 km/hr = 514.6kg km/hr
Skater (2) 55 kg* 0.0 km/hr = 0 kg km/hr
Total: 514.6

Eastward momentum (after collision):
Skater Pair: [514.6 kg km/hr] / [83 kg + 55 kg] = 3.72 km/hr north

Nortward momentum (before collision):
Skater (1) 83 kg* 0 km/hr = 0 kg km/hr
Skater (2) 55 kg* 7.8 km/hr = 429 kg km/hr

Northward momentum (after collision):
Skater Pair: [429 kg km/hr] / [83 kg + 55 kg] = 3.1087 km/hr east

Because north and east are at right angles to each other we can use the Pythagorean Theorem to solve for total velocity.

(3.1087)^2 + (3.72)^2 = 23.67
Square root of 24.023.67 is 4.2 So the velocity after impact is 4.2 km/hr

To find the direction we use the tangent.

Tan A = (east/north)
Tan A = (3.1087/3.72)
Tan A = (.82898)
A = 39.65degrees

I this system north is 0 deg, and east is 90.

So the final answer is that the pair move off at 4.2 km/hr at a heading of 39.65 degrees.

genius_generous answered 4 months ago
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