A 2.50 k g stone is sliding to the right on a frictionless horizontal surface at

Question

A 2.50kg stone is sliding to the right on a frictionless horizontal surface at 4.50m/s when it is suddenly struck by an object that exerts a large horizontal force on it for a short period of time. The graph in the figure shows the magnitude of this force as a function of time.

Part(a) What impulse does this force exert on the stone?

Part(b) Just after the force stops acting, find the magnitude of the stone's velocity if the force acts to the right.

Part(d) Just after the force stops acting, find the magnitude of the stone's velocity if the force acts to the left.

A 2.50kg stone is sliding to the right on a frictionless horizontal surface at 4.50m/s when it is suddenly struck by an object that exerts a large horizontal force on it for a short period of time. The graph in the figure shows the magnitude of this force as a function of time. What impulse does this force exert on the stone? Just after the force stops acting, find the magnitude of the stone's velocity if the force acts to the right. Just after the force stops acting, find the magnitude of the stone's velocity if the force acts to the left.

Explanation / Answer

Impulse = Area of curve = 2.5 * 1000 * 1 * 10^-3 = 2.5 kg.m/s

Velocity(right) = [(2.5*4.5) + (2.5)]/2.5 = 5.5 m/s

Velocity(Left) = [(2.5*4.5) - (2.5)]/2.5 = 3.5 m/s

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