A 4 kg block on a frictionless horizontal track is moving to the right at 8 m/s
ID: 1435546 • Letter: A
Question
A 4 kg block on a frictionless horizontal track is moving to the right at 8 m/s toward a stationary 6 kg block that is attached to a spring with k= 25 N/m. At t=0 the 4 kg block collides with the 6 kg block and the 2 blocks stick together (and stay together after the collision). At the moment of the collision the spring is not compressed at all. a) What is the kinetic energy and the momentum of the blocks before the collision? b) What is the kinetic energy and the momentum of the blocks just after the collision? c) Describe in words qualitatively what happens after t =0 (for the next 10 seconds or so). What do you expect to see? Now describe that mathematically with a graph showing x(t) and an equation for x(t). Use your intuition combined with what you have learned. d) At what time is x(t) largest? What is its magnitude at that time? d) Starting at t=0, plot the potential energy stored in the spring as a function of time. Please put labels and scales on both axes so that one can clearly see the maximum and minimum amounts of energy stored in the spring and the times at which those occur.Explanation / Answer
It is the example of completely inelastic collision.
In this type of collision, only linear momentum is conserved and velocity after the collision of both body is same.
Given Values : m1 = 4 kg, m2= 6 kg, k= 25 N/m, u1 = initial velocity of first body = 8 m/s
K.E. = (1/2)m1(u1)2 + (1/2)m2(u2)2…………….. equation 1
But u2 = 0, since second body is stationary
So Kinetic energy of first body is,
K.E.(Firsi body) = (1/2)m1(u1)2 = (1/2)*4*(8)2 = 128 J
K.E.(second body) = 0
Similarly momentum is,
P = m1(u1) + m2(u2)……………….. equation 2
Again u2 = 0 so momentum for first and second blocks before collision is,
P (First block)= m1(u1) = 4*(8) = 32 kg.(m/s)
P (Second block)= 0
K.E. = (1/2)(m1 + m2)(v)2……………..equation 3
As in completely inelastic collision, the final velocity of both the blocks is same, because they stick togrtherand move.
Here v is unknown quantity so we can calculate it by using formula,
V =(m1/m1 +m2) u1
V =(4/10)*8 = 3.2 m/s
Put this value in equation 3,
Total K.E. =
= 51.2 J
K.E. for first body = (1/2)* 4* (3.2)2 = 10.24*2 =20.48 J
K.E. for second body = (1/2)* 6* (3.2)2 = 10.24*3 = 30.72 J
c)
Describe in words qualitatively what happens after t =0 (for the next 10 seconds or so)
After t = 0, first block will collide with second block and stick together. Most important observation in this case is that blocks are stopped. Spring will not compressed initially.
After few seconds, both the blocks move together towards the mean position.
When blocks starts moving to and fro motion then they will pass the mean position twice in one oscillation, at that time when they reach the fixed point or maximum distance point in opposite direction, then only we can say that x(t ) is largest