H hi+ 2 (12 points, suggested time 25 minutes) A student releases a block of mas

Question

H hi+ 2 (12 points, suggested time 25 minutes) A student releases a block of mass m from rest at the top of a slide of height h. The block moves down the slide and off the end of Friction and air resistance are negligible. The overall height H of the setup is determined by the height of the room. Therefore, if h, is increased, h2 must decrease by the same amount so that the sum h+h2 remains equal lo I. The student wants to adjust hy and hy to make d as large as possible a table of height h, landing on the floor a horizontal distance d from the edge of the table. i. Without using equations, explain why making hy very small would cause d to be small, even though h2 would be large.

Explanation / Answer

(a) (i) Since the total height of the setup is the height of the room, therefore, if h1 is small and h2 is large as compared to h1 , then both will be equal to the total height of the room, therefore, since h1 is very small and h2 is very large as compared to h1 , then the horizontal landing distance d would be small since the small height h1 will decrease in the potential energy of the block therefore making it more kinetic due to the large height h2 ,therefore, when the block will move off the end of the table having large height h2 the will land up to a distance d0 which is less than d, the original horizontal landing distance.

(ii) Similarly, in this case the height of h1 which is larger than that of h2 which is very small , so from the larger height h1 , when the block will be released it will have more potential energy than the kinetic energy, therefore, the small height h2 of the will decrease the potential energy making it more kinetic, therefore, it will also land up to a distance d1 which is less than the original horizontal distance d.

(b) Taking the original situation given in the problem:

A block of mass m sliding from the topmost point of the setup having kinetic energy mgh1 and when it comes down to the table of height h2 from the ground it's potential energy will be mgh2 and after getting off from the table of height h2 it will land to the ground at a horizontal distance d having maximum kinetic energy equal to (1/2)md`2 , where d` is the velocity which is change of the distance per unit time, therefore the equation will be

mgh1 + mgh2 = (1/2)md`2

mg( h1 + h2) = (1/2)md`2

g( h1 + h2) = (1/2)d`2

d`2 = 2g ( h1 + h2)

d` = [ 2g( h1 + h2) ]1/2,   d` = d(d)/dt

d(d) = [ 2g( H ) ] 1/2 dt , H = h1 + h2

integrating both sides, we get,

d = t [ 2g(H) ]1/2 , which is the reduced form of the required equation.

(c) (i) For part a(i), where h1 is very small and h2 is large as compared to h1, therefore, we can say that h2 >> h1, but we cannot neglect h1 since they both will contribute to the total height H of the room which is constant, therefore, since we have concluded that the landing horizontal distance of the block will be d0 which is less than d , therefore as discussed in part a(i) the potential energy of the block at a height h1 will decrease since h1 is small , therefore taking the changed heights for this case h1` and h2` and h1` + h2` = H, the new potential energy will be mgh1`, similarly the new potential energy at a height h2` will be mgh2` and the new kinetic energy for the distance d0 will be 1/2 md0`2 where d0` is the derivtive of distance with respect to time which is velocity, therefore the equation will be

mgh1` + mgh2` = 1/2 md0`2

d0`2 = 2g ( h1` + h2`)

d0` = [2g (H)]1/2

d(d0) = [2g(H)]1/2 dt

integrating both sides we get,

d0 = t[2g (H) ]1/2 which is the required equation

(ii) Since in this case h1 is larger than h2 , therefore , the following the similar steps as done in part(i) , we get,

d1 = t [ 2g(H) ]1/2

(d)  If this experiment is to be performed on the moon then, since the mass of the moon is less than that of the earth, the gravitational force between two masses will be very less as that compared to the earth, therefore, if this experiment is performed in this condition the distance of the block landing to the ground will be greater than than of the distance when the same experiment is performed on the surface of the earth because the kinetic and potential energies of the block will be much greater than that of the kinetic and potential energies of the block when this experiment is performed on the surface of the earth.

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