A 145g object moving horizontally in the positive x direction with speed 30.0m/s

Question

A 145g object moving horizontally in the positive x direction with speed 30.0m/s, strikes a barrier at a 45 degree angle and rebounds horizontally in the postive y direction with unchanged speed. The object strikes the barrier at the origin.

a. Draw a creful figure showing the situation

b. Write doen the objects change in momentum using unit vector notation.

c. What is the magnitude of the impulse exerted by the barrier on the object?

d. What is the direction of the force exerted by the barrier on the object? give your answer as an angle measured counterclockwiise from the positive x axis.

Please explain fully to be rated!

Explanation / Answer

In the x direction: initial momentum = mv = 0.145 * 30 = 4.35 kgm/s

For final momentum = 0

change in x direction = 0 - 4.35 = -4.35

Similarly, y direction initialis zero final is 4.35 so

change in y direction = + 4.35

Then Using unit vector notation p = ( -4.35 i + 4.35 j ) kg-m/s

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