A boy jumps from a wall 3m high. What is an estimate of the change in momentum of the boy when he lands without rebounding?

A 5x10∧0 kgms-1

B. 5x10∧1 kgms-1

C. 5x10∧2 kgms-1

D. 5x10∧3 kgms-1

Answer: option C. 5 * 10² kg m/s

Explanation:

1) Change in momentum = momentum after landing - momentum just when landing

2) Momentum after landing = mass * velocity after landing = mass * 0 = 0

3) Momentum just when landing = mass * velocity when landing

4) Use the conservation of energy law to find the kinetic energy when landing, KE

KInetic energy when landing = potential energy when jumping

KE = PE

PE = mass*gh

g ≈ 10 m/s²; h = 3 m ⇒ PE = mass*10m/s²*3m = 30mass m²/s²

KE = [1/2]mass*v²

⇒ [1/2]mass*v² = 30mass m²/s²

⇒ v² = 60m²/s²

5) Momentum just when landing

P = mass * √60 m/s

P = mass * √60 m/s = mass * 7.75 m/s

6) To find the momentum you need the mass of the boy.

Assume different masses:

m = 6.45 kg ⇒ P = 6.45 kg * 7.75 m/s ≈ 50 kg m/s

m = 64.5 kg⇒ P = 64.5 kg * 7.75 m/s ≈ 500 kg m/s = 5 * 10² kg m/s

7) Conclusion: being 64.5 kg a reasonalbe mass for a boy, the answer is the option C. 5 * 10² kg m/s

Velocity of boy with which it hits the ground is

v = √ (2gH)

= √ (2 * 10 m/s^2 * 3 m)

= √ (60) m/s

Boy will not rebound.

So,

change in momentum = mv - 0

= Mass of boy * √ (60) m/s

So the answer would be C - best estimate.