# momentum in two-dimension

1. Marble 1 of mass 0.02 kg moving at a speed of 3 ms^-1 hits marble 2 at rest. Marble 1 bounces off with speed 2 ms^-1 in a direction making 45° with the initial direction. The collision is elastic. (a) Find the momentum given to marble 2.(b) Find the energy given to marble 2.(c) Hence find the mass of marble 2.

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a) Initial momentum of marble 1 = 0.06 kg m/s in horizontal direction (assumed)

Final momentum of marble 1 = 0.04 kg m/s in direction deviated 45 deg. from the horizontal direction.

So for the final momentum of marble 1, its component along the horizontal direction = 0.04 cos 45 = 0.02√2 kg m/s

Similarly, its component along the vertical direction = 0.02√2 kg m/s

Therefore marble 2's momentum has components:

In horizontal direction = (0.06 - 0.02√2) kg m/s

In vertical direction = 0.02√2 kg m/s

So the magnitude is 0.0423 kg m/s

b) Initial k.e. of marble 1 = 0.5 x 0.02 x 32 = 0.09 J

Final k.e. of marble 1 = 0.5 x 0.02 x 22 = 0.04 J

Since the collision is elastic, energy given to marble 2 = 0.09 - 0.04 = 0.05 J

c) Suppose that marble 2 has a mass m and gains velocity v after collision, we have:

mv = 0.0423 and mv2/2 = 0.05

From mv = 0.0423, we have m2v2 = 0.00181

(m2v2)/(mv2/2) = 0.0361

2m = 0.0361

m = 0.018 kg

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