P 19 Two blocks are released from rest from different heights in the apparatus s

Question

P 19 Two blocks are released from rest from different heights in the apparatus shown to the right. One of the blocks has four times the mass of the other block, and one block starts at collision four times the height of the other block (see diagram). The occurs blocks are released at different times so that they are timed here perfectly to collide at the exact bottom of the apparatus. When they collide they stick together and thereafter they move as a single unit. a. Find the height to which the combined blocks rise when they come to rest for the first time, and indicate which side of the apparatus (left, right, or center, if they stop when they collide) where this occurs. Provide this height as a number multiplied by h. b. Find the fraction of the total energy that the system starts with that is converted into thermal energy.

Explanation / Answer

(a) Consider here, m1 = 4m mass

and, m2 = m

Now, for determining the velocity of this mass at the collision point,

(1/2)*m1*v1^2 = m1*g*h

=> v1 = sqrt(2gh) towards right.(+ve sign consider)

for the second mass, m2 -

(1/2)*m2*v2^2 = m2*g*4h

=> v2 = sqrt(8gh) (-ve sign consider)

Now suppose the velocity of the combined mass is v.

apply conservation of momentum -

m1v1 + m2v2 = (m1+m2)v

=> 4m*sqrt(2gh) - m*sqrt(8gh) = (4m+m)*v

=> 4*sqrt(2gh) - 2*sqrt(2gh) = 5v

=> 2*sqrt(2gh) = 5v

=> v = 0.4*sqrt(2gh)

Now suppose h1 is the height attained by the combined mass in the right direction -

so -

(1/2)*(m1+m2)*v^2 = (m1+m2)*g*h1

=> h1 = v^2/2g = [0.4*sqrt(2gh)]^2 / 2g = 0.16*(h)

(b) Total initial KE energy, KE1 = 4mgh + mg(4h) = 8mgh

Final KE, KE2 = (1/2)*(m1+m2)*v^2 = 0.5*5m*[0.4*sqrt(2gh)]^2 = 2.5m*0.16*2gh = 0.8mgh

Loss in KE = Energy converted into thermal energy = 8mgh - 0.8mgh = 7.2mgh

So, the fraction of energy that is converted into thermal energy = (7.2mgh) / (8mgh) = 0.90.

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