Conservation of Momentum help please? An 80-kg QB jumps straight up before throwing a 0.43-kg football...?

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 [(VQx)f] will he be moving backward just after releasing the ball?

I'm having trouble understanding conservation of momentum problems, and any help--like which formula to start with--would be really helpful. Thank you!

Answer 1

Hello Liseuti, any way friends have solved them in the right way.

Let me explain the theory behind and give you the formula with which in future you yourself can solve problems like this.

Law of Conservation of momentum states that the momentum of a system in a particular direction remains constant unless it is disturbed by any external force

If u1 u2 are the initial velocities say along X direction of masses m1 and m2 respectively, and v1 and v2 are the velocities after some collision among the two masses, then the momentum before collision ie m1u1 + m2 u2 will be equal to the sum of the final momentum ie m1v1 + m2v2

Hence using m1u1 + m2 u2 = m1v1 + m2 v2 you can solve for the unknown very easily. All the best.

Answer 2

My favorite QB would have to be Jim McMahon. He's the only QB to ever lead the Bears to a Super Bowl win. It's hard to pick a favorite member, but I've have to say TJ. He makes me laugh and he's a true Bengals fan.

Answer 3

Horizontal momentum before the throw is equal to 0, because nothing is moving horizontally, so the momentum afterwards has to equal 0.

0=.43kg(15m/s)+(80kg)(v)

solve for v to find that v=-.08m/s (the negative tells you that it is in the opposite direction of the football)

Answer 4

(VQx)f = 0.081 m/s