You are riding a skateboard and you are traveling at a speed v = 2.0 m/s. You ha

Question

You are riding a skateboard and you are traveling at a speed v = 2.0 m/s. You have a heavy ball, m = 3.0 kg in your hand which you now throw backwards. Just after the ball is released, it is measured to be traveling at a horizontal speed of u = 1.0 m/s with respect to ground. Suppose the total mass of you, the skate board and the ball is M = 60 kg, then

(a) will you be traveling faster or slower after you throw the ball?

(b) Let’s now make things quantitative. What is the initial total momentum of the system (you + board + ball) before the ball is released?

(c) What is the momentum of (you + board) just after the ball is released?

(d) What then must be your final speed, vf, after the ball is released?

(e) Now do the whole problem using symbols, instead of numbers. So, find an expression for vf in terms of v, u, M, and m.

Explanation / Answer

(a) the person will be travelling faster after the ball is thrown.

(b)initial total momentum of the system (you + board + ball) before the ball is released = M V

Pi = 60(2) = 120 kgm/s

(c)since momentem before= momentum after

so  momentum of (you + board) just after the ball is released = 120 kgm/s

(d) momentem before= momentum after

M V =mu +( M-m)Vf

60(2)= 3(-1)+ (60-3) Vf

120=-3 +57Vf

(e)

momentem before= momentum after

M V =m(-u) +( M-m)Vf

Vf = MV+mu/(M-m)

123= 57Vf

Vf= 123/57 =2.16m/s

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