You can jump 3 m (10 feet) off of a diving board and into a swimming pool and be uninjured. However, if you jumped from the same height onto concrete it would be very bad. Use the principles of impulse and momentum to explain this.

2) A 125 kg linebacker running to the right at 2.5 m/s and an 80 kg quarterback running to the left at 3 m/s collide in midair. The linebacker holds on to the quarterback and they move off together. What is their velocity (magnitude and direction) after this collision?

1) this is due to force due to momentum change of a body falling from such a height.

2) they move with velocity of 2.69 m/s towards the right.

Explanation:

1) in the case where one jumps 3 meters into water, the water deform on impact extending the time of impact for your body's momentum to get to zero, so that the time it takes for the momentum of your body due to your mass and your velocity to get to zero is relatively long lessening the force felt due to the impact.

For a fall from the same height onto a concrete, the concrete does not deform at all. This means that there is a very short time for your momentum to get to zero and hence the force is huge. In order to cushion this a little, your body deforms in a bid to reduce the force and this is very deadly.

This is in line with Newton's second law of motion which states that the rate of change of momentum is proportional to the force produced.

2)

M1 = 125 kg

M2 = 80 kg

V1 = 2.5 m/s

V2 = 3 m/s

From momentum conservation, total initial momentum must be equal to the total momentum after collision.

Momentum = mass x velocity.

That is,

M1V1 + M2V2 = (M1 + M2) Vf

Where Vf is their final velocity together.

Note that the masses stick together because the collision is inelastic.

Substituting and solving, we have

(125 x 2.5) + (80 x 3) = (125 + 80) Vf

312.5 + 240 = 205Vf

552.5 = 205Vf

Vf = 552.5/205 = 2.69 m/s

Note that the linebacker runner has the greater momentum 125 x 2.5 = 312.15

This means they both move to the right at velocity of 2.69 m/s