A 0.159 kg glider is moving to the right on a frictionless, horizontal air track

Question

A 0.159 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.860 m/s . It has a head-on collision with a 0.298 kg glider that is moving to the left with a speed of 2.30 m/s . Suppose the collision is elastic.

Find the magnitude of the final velocity of the 0.159 kg glider. (equal to -3.26 m/s)

Find the magnitude of the final velocity of the 0.298 kg glider.

Find the direction of the final velocity of the 0.298 kg glider.

Having trouble performing the equations in order to get the value for slider of .298kg.

Explanation / Answer

here m1 = 0.159kg m2 = 0.298kg v1 = 0.860m/s v2 = -2.30m/s (since it is from left)

Use the following equations for elastic collisions, which can be derived by combining the conservation of energy and conservation of momentum laws. The prime ' stands for velocity after the collision.

v'1 = [(m1 m2)/(m1 + m2)]v1 + [2m2/(m1 + m2)]v2

v'2 = [2m1/(m1 + m2)]v1 + [(m2 m1)/(m1 + m2)]v2

v'1 = [(0.159-0.298) / (0.159+0.298)]0.860 +[ 2*0.298 / (0.459)](-2.30) => -3.26m/s

Here the second equation is the one we want, since it represents the speed for the second glider after the collision.

v'2 = [2*0.159/(0.159 + 0.298)]0.860 + [(0.298 0.159)/(0.159 + 0.298)](2.30) = -0.100 m/s

The magnitude of the final velocity of the 0.298 kg glider is 0.100 m/s

any motion towards the left is arbitrarily negative and motion to the right, positive hence 0.298kg glider is towards left

Hope this will helps you

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