[2-d collision on a frozen pond] while doing physics experiments on a frozen pond's frictionless horizontal surface, sam, of mass 80.0 kg, is given a push and slides eastward. abigail, of mass 50.0 kg, is sent sliding northwards. they collide (whump!), and afterwards sam is moving at 37.0° north of east at 6.00 m/s, while abigail is moving 23.0° south of east at 9.00 m/s.

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14-10-2022
Physics

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[2-d collision on a frozen pond] while doing physics experiments on a frozen pond's frictionless horizontal surface, sam, of mass 80.0 kg, is given a push and slides eastward. abigail, of mass 50.0 kg, is sent sliding northwards. they collide (whump!), and afterwards sam is moving at 37.0° north of east at 6.00 m/s, while abigail is moving 23.0° south of east at 9.00 m/s.

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charithran8975 charithran8975 · Answer 1 · 2022-10-18T10:24:32-04:00

We could calculate the initial and final kinetic energy after computing the velocities prior to the impact. So, the total kinetic energy of the two people decrease during the collision by -638.73 J.

Since the velocity before to the collision is unknown, it is impossible to calculate the kinetic energy. So, we'll start by computing the velocities before to the impact.

If all external forces have a zero vector total, momentum is said to be conserved. In other terms, it may be said that the system's momentum prior to the collision and its after momentum should be identical.

This indicates that even if we know little or nothing about the internal forces of the system, we can still calculate the system's momentum.

The kinetic energy prior to and following the collision must be calculated in order to calculate the change in kinetic energy.

Due to the unknown velocity before to the contact, it is impossible to calculate the kinetic energy. As a result, we'll start by computing the velocities that existed before the impact.

The first step is to calculate the x and y components of the velocity following the collision. Sam might be represented by variable S, and Abigail by variable A.

v S = v S x + v S y = 6.00 m / s cos 37 + 6.00 m / s sin 37 v S = ( 4.79 I + 3.61 j ) m / s v A = v A x + v A y = 9.00 m/s cos 23 + 9.00 m/s sin 23 v A = ( 8.28 I + 3.52 j ) m / s

Then, keep in mind the equation for momentum conservation: p 1 = p 2

where p 1 represents the momentum prior to the collision and p 2 represents the momentum following the collision. Since Sam is sliding eastward before to the collision in the aforementioned scenario, we will only utilize the x component of the velocity after the collision of Sam and Abigail to compute the velocity of Sam before the collision, and we will also set Abigail's momentum to zero.

We could now calculate the initial and final kinetic energy after computing the velocities prior to the impact.

KE1=12mS(vS1)

2+12mA(vA1)

2=12(80 kg)(9.97 m/s)

2+12(50 kg)(2.26 m/s)

2KE1=4103.73 J

KE2=12mS(vS1)

2+12mA(vA1)

2=12(80 kg)(6.00 m/s)

2+12(50 kg)(9.00 m/s)

2\sKE2=3465 J

The difference between the end and starting kinetic energy could be used to calculate the change in kinetic energies.

KE= -638.73J

Complete question is:

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with mass 80.0 kg, is given a push and slides eastward. Abigail, with mass 50.0 kg, is sent sliding northward. They collide and after the collision, Sam is moving at 37.0 degrees north of east with a speed of 6.00m/s and Abigail is moving at an angle 23.0 degrees south of east with a speed of 9.00 m/s. By how much did the total kinetic energy of the two people decrease during the collision?

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