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2 years ago

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A 2.40 kg stone is sliding in the +x-direction on a horizontal, frictionless surface. It collides with a 4.00 kg stone at rest.

After the collision the 2.40 kg stone is moving at 3.60 m/s at an angle of 30.0∘ measured from the +x-direction toward the +y-direction and the 4.00 kg stone is moving at an angle of 45.0∘ measured from the +x-direction toward the −y-direction. What is the y-component of momentum of the 2.40 kg stone after the collision?

1 answer:

tester [92]2 years ago

5 0

Answer:

4.32 kg.m/s

Explanation:

SInce the 2.4 kg stone is mobing at speed of 3.6m/s at an angle of 30 degree measured from the x+ direction toward y+ direction. Its y-component momentum is product of its velocity in the y-direction and its mass

$P_y = m*v_y = m*v*sin(\theta)$

$P_y = 2.4*3.6*sin(30)$

$P_y = 2.4*3.6*0.5 = 4.32 kg.m/s$

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