The two masses (2.2 kg and 3.6 kg) shown in the figure are each initially 1.8 m
ID: 1456844 • Letter: T
Question
The two masses (2.2 kg and 3.6 kg) shown in the figure are each initially 1.8 m above the ground, and the massless frictionless pulley is 4.8 m above the ground. Ignore the mass of the cord, (a) What is the acceleration of the lighter mass at the moment the heavier mass hits the ground? (b) What is the velocity of the lighter mass at the moment the heavier mass hits the ground? (c) What maximum height does the lighter mass reach after the system is released? Assume the mass doesn't hit the pulley.Explanation / Answer
(a) Acceleration of the lighter mass in the upward direction shall be constant until the heavier mass strikes the surface.
The acceleration is given by:
a = g*[(M - m)/(M+m)] = 9.81*[(3.6-2.20)/(3.6+2.2)] = 9.81* 1.4/5.8 = 2.37 m/s^2
(b) Velocity of the lighter mass, when the heavier mass hits the ground.
In this case the distance travelled by the lighter mass is 1.8 m
So, applying the formula:
v^2 = u^2 + 2*a*d
=> v^2 = 0 + 2*2.37*1.8
=> v^2 = 8.532
=> v = 2.92 m/s
(c) Here, we have the launch speed of the lighter mass after travelling a distance of 1.8 m, v = 2.92 m/s
dy after launch speed reached: y2 = v²/2g = 2.92^2/(2*9.81) = 0.434 m
Hmax = 1.80 + 0.434 = 2.234 m
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.