A 0.500 kg block is released from rest at the top of a frictionless track 2.50 m

Question

A 0.500 kg block is released from rest at the top of a frictionless track 2.50 m above the top of a table. It then collides elastically with a 1.00 kg object that is initially at rest on the table.
a. Determine the velocities of the two objects just after the collision. (no magic formulas - show all of your work)
b. How high up the track does the 0.500 kg object travel back after the collision?
c. How far away from the bottom of the table does the 1.00 kg object land, given that the table is 2.00 m high?
d. How far away from the bottom of the table does the 0.500 kg object eventually land?

Explanation / Answer

a)as we know P.E=K.E.

mxgxh = 1/2xmxu1^2

so,

u1= sqrt (2xgxh) = 7 m/s and u2 = 0 (sytem strt frm rest)

so,

v1 = (m1 - m2 ) x u1/(m1 + m2 ) = - 2.4 m/ s

v2 = (2 x m1 x u1/(m1 + m2 ) = 4.5m/s

b)now by above formulae

h = v1^2 / 2g= 0.28 m

c) & we know time = ( 2h /g)^(1/2) = 0.70sec

x 2= v2 x t = 2.9 m

d) x1 = v1 x t = 1.5 m

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