Guys please help me out i have an exam in an hour!! A ball of mass M is dropped
ID: 1265113 • Letter: G
Question
Guys please help me out i have an exam in an hour!!A ball of mass M is dropped by 2m in a fall experimint.
A-what is the speed of the ball after the 2m drop?
B- it now hits anouther ball of mass 3M that is at rest sispended on tissue paper ( the tissue will break without any extra force). The two balls collide with a perfect elastic collision. What is the forward speed of the second ball immeditly after the collision ?
C- to what height does the first ball rebounded up again after the collision ? Guys please help me out i have an exam in an hour!!
A ball of mass M is dropped by 2m in a fall experimint.
A-what is the speed of the ball after the 2m drop?
B- it now hits anouther ball of mass 3M that is at rest sispended on tissue paper ( the tissue will break without any extra force). The two balls collide with a perfect elastic collision. What is the forward speed of the second ball immeditly after the collision ?
C- to what height does the first ball rebounded up again after the collision ? Guys please help me out i have an exam in an hour!!
A-what is the speed of the ball after the 2m drop?
B- it now hits anouther ball of mass 3M that is at rest sispended on tissue paper ( the tissue will break without any extra force). The two balls collide with a perfect elastic collision. What is the forward speed of the second ball immeditly after the collision ?
C- to what height does the first ball rebounded up again after the collision ?
Explanation / Answer
A)tThe speed just before hitting the floor is simply
sqrt[2gh] where h is now 2m, so v before impact is
v=sqr(2gh)
=sqrt(2*9.8*2m)
V=6.2sqrt(m) m/sec----Ans
b)Using law of conservation
v2' = (m1v1 + m2v2)/(m1+m2)
=(Mv1+3Mv2)/(M+3M)
=(Mv1+3Mv2)/4M
v2'=(v1+3v2)/4
c) mgh = 1/2mv^2 (masses cancel out )_
(9.8)(h)= (1/2)v2^2
h= 0.05v2^2 m
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