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

Question

A 2.50 kg stone is sliding to the right on a frictionless horizontal surface at 6.00 m/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 (Figure 1) 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 direction 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.

Just after the force stops acting, find the direction of the stone's velocity if the force acts to the left.

Explanation / Answer

a) impulse = F dt = area of the F-t Curve

impulse = 2.5*10^3*(16-15)*10^-3 = 2.5 kgm/s

b) change in momentum = final momentum - initial momentum

2.5 = 2.5(V-6)

V = 7 m/s

the velocity is 7 m/s

c) the direction will be the same as the initial direction of motion of the stone

d) when the force acts to the left

change in momentum = initial momentum - final momentum

2.5 = 2.5( 6 - V)

V = 1 m/s

e) the direction will be the same as the initial direction

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