Question

An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?
Model

The interaction at study in this problem is the action of throwing the ball, performed by the quarterback while being off the ground. To apply conservation of momentum to this interaction, you will need to clearly define a system that is isolated or within which the impulse approximation can be applied.

Which of the following best describes why you can analyze the interaction described in this problem using conservation of momentum?

(A) The throwing action is quick enough that external forces may be ignored.
(B) There are no external forces acting on the system.
(C) External forces don't act on the system during the jump.
(D) Conservation of momentum is always the best way to analyze motion.

Answer 1

Answer:

Explanation:

We shall apply law of conservation of momentum .

m₁ v₁ = m₂ v₂

m₁ = 80 kg

v₁ = ?

m₂ = .43 kg

v₂ = 15 m /s

80 v₁ = .43 x 15

v₁ =  .43 x 15  / 80

= .08 m /s

8 cm /s

option A is correct.

Law of conservation of momentum is applied only where no external force acts . Here during the period when ball was thrown , apart from internal force , external force like gravitational force was also acting but we still applied this law because the time duration of action of throw was very small

so we neglected the effect of external force . So it was just an approximation.