Question

Two friends, Al and Jo, have a combined mass of 166 kg. At the ice skating rink, they stand close together on skates, at rest and facing each other. Using their arms, they push on each other for 1 second and move off in opposite directions. Al moves off with a speed of 5.9 m/sec in one direction and Jo moves off with a speed of 9.7 m/sec in the other. You can assume friction is negligible.

1)What is Al's mass? kg

2)What is Jo's mass? kg

3)If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Al on Jo? N

4)If you assume the force is constant during the 1 second they are pushing on each other, what is the magnitude of the force of Jo on Al? N

Answer 1

1.

Let m1 be mass of Al and m2 be mass of Jo

m1+m2 = 166 so m2 =166- m1

According to law of conservation of momentum, the sum of initialmomentum =sum of final momentum

Sum of initial momentum = 0 ( both were at rest )

So sum of final momentum = 0

m1* 5.9 + m2 * -9.7 = 0

m1* 5.9 + (166-m1) *-9.7 = 0

5.9m1 - 1610.2 + 9.7 m1 = 0

15.6 m1 - 1610.2 = 0 or m1= 1610.2/15.6 =

103.2 Kg.

So the mass of al = 103.2 Kg and

2.

mass of Jo = 166- mass of Al= 166-103.2 = 62.8 Kg

3.

We have force F = ma

Where m is mass and a is acceleration

Mass of jo = 62.8 Kg

Acceleration of jo a= final - initial velocity/time

a =-9.7 -0 /1 = -9.7 m/s²

So force of Al on Jo = ma = 62.8* -9.7 = -609 N

So the magnitde of force of Al on Jo = 609 N

4. We have the equation for force F = ma

Where m is mass and a is acceleration.

Acceleration of Al,

a = final velocity - initial velocity/time

= 5.9-0 / 1 = 5.9 m/s²

So the force F = ma = 103.2* 5.9 = 608.88 N

So the magnitde of force of Jo on Al is 609 N

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