Question

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 1

Answer: option C. 5 × 10² kg m/s

Step-by-step 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

Answer 2

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.