An object slides down frictionless slope from a certain height and into a vertic

Question

An object slides down frictionless slope from a certain height and into a vertical loop (like a roller coaster). If the vertical loop has a radius of 5 meters, what is the minimum height needed for the object to safety make it through the loop? After the object completes the vertical loop from part (a). the object is thrown into a 5O kg cart at level ground. If the mass of the object Is 75 kg. what is the resultant speed of the cart and the object together? The object in the cart from part (b.) is changed to 1S m/s after a certain duration of time. At this point, the cart carrying the object hits a barrier at the edge of the cliff which Immediately stops the cart, but throws the object forward off the edge of the cliff. How far away from the edge of the cliff will the object be thrown if the height of the cliff is 2S meters? If a cushioned mat was placed equal to the resultant distance from the edge of the cliff in part (c) as to catch the object, what amount of average force was applied to the object if the cushioned mat stops the object s fall

Explanation / Answer

a. think in terms of conservation of energy. Since there is no friction, and it starts with gravitational potential energy = mgh, it should get up to a maximum height of 'h' in other words > 5m b. presumably the exit point from the loop is at "ground level" or h = 0 so all of the gravitational potential we started with is now kinetic energy. ie mgh = .5mv^2 or v = sqrroot( 2gh) now you can use conservation of momentum, assuming that the cart was at rest, and the collision is perfectly inellastic (they stick together) m(obj)*v(from above) + m(cart)*0 = ( m(obj) + m(cart) ) v(final) v(final) = [m(obj) v(from above)]/ [ m(obj) + m(cart)] c) reverse of b 15m/s ( m(obj) + m(cart) = initial momentum final momentum is all transferred to the object so that (15m/s * (m(obj) + m(cart)) / m(obj) = velocity of the object off the cliff. calculate the time it takes to fall from x = 1/2gt^2 x is height of cliff, initial vertical velocity is0 then multiply the time it is in the air, by its initial velocity off the cliff to find how far it flew d. impulse is the average change in momentum. now the object has a momentum that results from horizontal and veritcal velocity v off cliff * m(obj) = momentum (horizontal) t(in air) * g * m(obj) = momentum (vertical) divide both components of the total momentum by the time of the impact to get the impulse (average force) but remember that it is the mattress acting on the object so the sign of the force should be opposite the sign of the incoming momentum vector. There ya go

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