Question

In: Physics

Two people are headed towards each other on skateboards and experience very little friction or airresistance....

Two people are headed towards each other on skateboards and experience very little friction or airresistance. One person has massm1= 40 kg and one has mass m2= 80 kg. If they are going equal velocities(v = 3 m/s)before the collision, which direction will they be going afterward? How fast?What if we allow the lighter person to start off on a hill of height h. What should h be so that after the collision both skateboarders have stopped?

Solutions

Expert Solution

Mass of skateboarder 1 = m1 = 40 kg

Mass of skateboarder 2 = m2 = 80 kg

The skateboarders are headed towards each other with equal velocities.

Let us consider that the velocity of skateboarder 1 is to the right side and take right side as the positive direction, hence velocity of skateboarder 2 is to the left side and will be negative.

Initial velocity of skateboarder 1 = V1 = 3 m/s

Initial velocity of skateboarder 2 = V2 = -3 m/s

After collision both skateboarders will move as a single unit with same velocity.

Velocity after collision = V3

By linear momentum conservation,

m1V1 + m2V2 = (m1 + m2)V3

40(3) + 80(-3) = (40 + 80)V3

V3 = -1 m/s

The negative sign indicates that the skateboarders will be going to the left side after the collision that is the same direction as the initial velocity direction of skateboarder 2.

The skateboarders after collision will be going in the direction initially the skateboarder 2(heavier person) was going in with a velocity of 1m/s.

Now we allow the lighter person(skateboarder 1) to start off on a hill of height h, and after the collision both the skateboarders stop.

Velocity of skateboarder 1 before collision = V4

Velocity of skateboarder 2 before collision = V2 = -3 m/s

For the skateboarders to stop after the collision the momentum before the collision should be zero.

m1V4 + m2V2 = 0

40(V4) + 80(-3) = 0

V4 = 6 m/s

Potential energy of the skateboarder at height h gets converted into kinetic energy as the skateboarder goes downhill

PE = KE

mgh = m(V42)/2

40 x 9.81 x h = 40(62)/2

h = 1.83 m

The height h so that after the collision both skateboarders stop is 1.83 m


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