Two balls are connected by a string that stretches over a massless, frictionless pulley. Ball 1 has a mass of 0.45 kg and is held 0.74 m above the ground. Ball 2 has a mass of 5.7 kg and is held 0.88 m above the ground. When the balls are released, ball 2 falls to the ground, looses 30 % of its kinetic energy and rebounds to some maximum rebound height. When the balls are released, ball 1 travels to some maximum height before starting to fall. Assume that ball 1 reaches its maximum height during ball 2's rebound so that the string doesn't pull.
Calculate the maximum height of ball 1 from and ground and the rebound height of ball 2.
2. Relevant equations
3. The attempt at a solution
can you check my work i am stuck
I would calculate the kinetic energy of ball 2 just before and after it hits the ground, using
Just before impact, ball 2 (M) and ball 1 (m) have acquired kinetic energy
(1/2)mV^2 + (1/2)MV^2
= M g*0.74 - m g*0.74 Solve for V of both balls, and the kinetic energy of ball 2.
(1/2)(m + M)V^2 = 0.74 (M-m)g
V^2 = [(M-m)/(M+m)]g*1.48 = 12.38 m^2/s^2
V = 3.52 m/s
= After impact, ball 2 will have 70% of the pre-impact kinetic energy, and its velocity will be V' = sqrt(0.7)V= 0.8367V = 2.94 m/s
Right after ball 2's impact, Ball 1 will continue to rise for a while because it suffered no impact and maintained its velocity V when ball 2 hits the ground. There will be no tension in the string while ball 2 and ba1l 1 both rise.
Ball 1 rises a distance H1 given by
M g H1 = (1/2) M V'^2 = (1/2)M(0.7)V^2
H1 = (0.7)V^2/(2g) = 0.35 V^2/g
When ball 1 hits the ground, ball 2 has already risen from 0.74 to 0.74 + 0.88 = 1.62 m above the ground. It will rise an additional distance until its kinetic energy (1/2) mV^2(at impact)is converted to potential energy
i am getting the wrong anwser
(1/2)mV^2 + (1/2)MV^2
= M g*0.74 - m g*0.74 Solve for V of both balls, and the kinetic energy of ball 2.
Shouldnt it be Mg*0.88 instead of Mg*0.74 for ball 2
2 answers
2 answers