Physics collision

Question

A rubber stopper collides inelastically with the ground. during the collision, the magnitude of the force exerted on the stopper is F(t)= 2x10^5t-3x10^7t^2 the collision last for 6.67 ms

calculate he average force exerted on the stopper

the stopper strikes the floor at 15m/s at an angle of -30 degrees with respect to it and leaves the floor at an angle of 60 degrees at 10 m/s. What is the mass of the stopper and the direction of the force exerted on the stopper by the floor

Answer 1

1.) you have force as function of time. Integrate wrt t from 0 to 6.67 ms. This gives you impulse. Then divide by total time (6.67 ms) and you have average force.

 

2.) from part 1 you will have impulse Which equals change in momentum. You can assume mass of stopper doesn't change so it's constant. You need to look at change in momentum in x and y coordinate.

F_xt = m(v_fx_v_ix)

similar in y direction.

 

you know both initial and final velocities in x and y direction. 

You can then use 

(F_x²+F_y²)^.5 = (px^2-py^2)^2/t with this you should be able to solve for m.

once you have m you can solve for the force in x and y direction and then you can determine the direction of the force 

tan(theta) = force in y divided by force in x

 

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