m1 = 0.450 kg block is released from rest at the top of a frictionless track h1
ID: 1982929 • Letter: M
Question
m1 = 0.450 kg block is released from rest at the top of a frictionless track h1 = 2.25 m above the top of a table. It then collides elastically with a 1.00 kg block that is initially at rest on the table.(a) Determine the velocities of the two blocks just after the collision.
volocity of m1
velocity of m2
(b) How high up the track does the 0.450 kg block travel back after the collision?
(c) How far away from the bottom of the table does the 1.00 kg block land, given that the table is 2.30 m high?
(d) How far away from the bottom of the table does the 0.450 kg block eventually land?
Explanation / Answer
Masses m 1 = 0.45 kg
m 2 = 1 kg
Initial height h 1 = 2.25 m
Initial velocity of m 1 is u 1 =[2gh 1]
= 6.64 m/ s
Initial velocity of m 2 is u 2 = 0
For eleastic collision , coefficient of restitution e = 1
( V - v ) / ( u 1- u 2) = 1
V - v = 6.64
V = v + 6.64 ------------( 1)
From law of conservation of momentum , m1u1 + m2u2 = m1 v + m 2 V
2.988 + 0 = 0.45 v + 1(v + 6.64)
= 1.45 v + 6.64
v = -2.51 m / s
From eq( 1) , V = 4.12 m / s
(b). Required height h2 = v 2 / 2g
= 0.3214 m
(c). Height of table H = 2.3 m
Time taken to reach the ground t = [2H / g ]
= 0.685 s
Required distance R = Vt
= 2.822 m
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