Two balls are connected to 60 - cm - long light strings and the other ends of th
ID: 2278368 • Letter: T
Question
Two balls are connected to 60 - cm - long light strings and the other ends of the strings are fixed together as shown in the figure. One of the balls has a mass of 2.0 kg and is raised up and to the right unitil it is 12.00 cm higher than the other ball, which has a mass of 3.0 kg. The upper ball is releaed from from rest and sticks to the lower ball when they collide. Find the
a) frequency
b) max angular dispacement
c) max height
d) max speed of the subsequent motion.
Two balls are connected to 60 - cm - long light strings and the other ends of the strings are fixed together as shown in the figure. One of the balls has a mass of 2.0 kg and is raised up and to the right until it is 12.00 cm higher than the other ball, which has a mass of 3.0 kg. The upper ball is released from from rest and sticks to the lower ball when they collide. Find the frequency max angular displacement max height max speed of the subsequent motion.Explanation / Answer
strings are attached to the same fixed point and this is a simple pendulum problem.
(a) The frequency of the pendulum does not depend on the mass of the bob, only the length L of the string :
f = (1/2pi) sqrt (g/L)
= (1/6.283) sqrt(9.8/0.12) = 1.44 Hz.
(b) The maximum angular displacement (or amplitude A) is given by
cosA = (60-1.92)/60 = 0.968
A = 0.2537 rad
= 14.5 degrees.
(c) The coalesced mass gradually loses its KE of (1/2)Mv^2 as it swings, and it reaches maximum height H where its PE of MgH is equal to its initial KE, ie
MgH = (1/2)Mv^2
2gH = v^2.
Using the above formulas we have
H/h = (v/u)^2
= (m/M)^2
= (2/5)^2
= 0.16.
So the maximum height of the coalesced mass is
H = 12cm x 0.16 = 1.92 cm.
(d) There will be some loss of kinetic energy when the balls collide, but momentum is conserved.
At the point of collision, the 2kg ball has lost potential energy PE of mgh where m=2kg, g=9.8m/s^2 and h=12cm=0.12m. It has gained an equal amount of KE of (1/2)mu^2 where u is the speed :
(1/2)mu^2 = mgh
u^2 = 2gh
= 2*9.8*0.12
= 2.353.
u = 1.534 m/s.
On collision, the balls coalesce and have mass M=5kg and speed v where
Mv = mu
because of conservation of momentum.
v = mu/M
= 2*1.534/5
= 0.613 m/s.
This is the maximum speed of the coalesced mass.