A block of mass m 1 = 2.1 kg initially moving to the right with a speed of 4.2 m
ID: 1340100 • Letter: A
Question
A block of mass m1 = 2.1 kg initially moving to the right with a speed of 4.2 m/s on a frictionless, horizontal track collides with a spring attached to a second block of mass m2 = 4.4 kg initially moving to the left with a speed of
(A) Find the velocities of the two blocks after the collision.
ANSWER -5.55 m/s
(B) During the collision, at the instant block 1 is moving to the right with a velocity of 1.0 m/s as in figure (b), determine the velocity of block 2.
ANSWER -1.47272727 m/s
(C) Determine the distance the spring is compressed at that instant.
ANSWER 0.34685 m
QUESTIONS--->
What if m1 is initially moving at 3.6 m/s while m2 is initially at rest?
(a) Find the maximum spring compression in this case.
X= _____?_____m
(b) What will be the individual velocities of the two masses (v1 and v2) after the spring extended fully again? (That is, when the two masses separate from each other after the collision is complete.)
V1 = _____?_____m/s to the left
V2= ______?_____m/s to the right
Explanation / Answer
a) m1 = 2.1 kg, m2 = 4.4 kg v1 = 3.6 m/s
KE = 0.5 x 2.1 x 3.6^2 = 13.6 J, this is transfered to the spring as PE
maximum compression will occur when the two blocks move with the same speed V
2.1 x 3.6 = (2.1 + 4.4) x V
V = 1.16 m/sec
KE of the center of mass = 0.5 x 6.5 x 1.16^2 = 4.37 J
balance of KE will be transfered to the spring
PE of the sprin = 0.5kx2 = (13.6 -4.37) , where k= 547N/m spring constant
x = 0.18 m , amximum compression of the spring,
b) m2 is at rest Total momentum before collision = 3.6x2.1 = 7.56 kg-m/s
= 2.1v1+4.4v2 after collision, where v1 and v2 are velocities of m1 and m2
2.1v1 +4.4V2 = 7.56 ------------(1)
we have velocity of seperation = velocity of approach
v2-v1 = 3.6 ---------------------(2)
solving 1 and 2 we get v1 and v2
v1 = -1.88 m/sec m1 travels in opposite direction after collision
v2 = 1.72 m/sec m2 travels forward
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