Two blocks of masses m 1 = 2.00 kg and m 2 = 4.15 kg are each released from rest

Question

Two blocks of massesm1= 2.00 kg andm2=4.15kg are each released from rest at a height ofh=5.50m on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision. (Let the positive direction point to the right. Indicate the direction with the sign of your answer.)

(a) Determine the velocity of each block just before the collision. v1i = m/s v2i = m/s

(b) Determine the velocity of each block immediately after the collision. v1f = m/s v2f = m/s

(c) Determine the maximum heights to which m1 and m2 rise after the collision. y1f = m y2f = m

Explanation / Answer

a) Using energy
v=sqrt(2*g*h)
v1i=v2i=sqrt(2*9.81*5.5)
10.39 m/s


b) Since the collision is elastic
momentum and energy are conserved

2*10.39 -4.15*10.39 =2*v1f+4.15*v2f

.5*2*10.39 ^2+.5*4.15*10.39 ^2=
.5*2*v1f^2+.5*4.15*v2f^2

you will get v1,v2

c)
Using energy
h=.5*v^2/g

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.