A body of mass 400 g is moving along a smooth surface at a velocity of 10 m s–1 east. It strikes a body of mass 650 g, initially at rest, and then the 400 g body moves at a velocity of 8 m s–1 in a direction E35°N. What is the velocity of the 650 g object?

v = 2.53 m/s at E 53.1° S

Explanation:

Conservation of momentum

The 400 g object has no North-South velocity, so the initial momentum in that direction is zero. The total momentum after collision must also be zero

0 = 400(8sin35) + 650vy

vy = -2.82376... m/s

In the East direction

400(10) = 400(8cos35) + 650vx

vx = 2.121097...m/s

v = √(2.12² + 2.82²) = 3.531667... ≈ 2.53 m/s

θ = arctan(vy/vx) = arctan(-2.82/2.12) = -53.087... ≈ E53.1°S

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