In the figure below, two carts are collided on a frictionless surface. Block 1 o
ID: 1341078 • Letter: I
Question
In the figure below, two carts are collided on a frictionless surface. Block 1 on the left has a mass of m_1 = 4.220 kg and block 2 on the right has a mass of m_2 = 21.10 kg. Block 1 is travelling to the right with a speed v_1 = 5.000 m/s and block 2 is travelling to the left with a speed v_2 = 2.500 m/s. Block 1 has an ideal, massless spring attached to it. Is this an elastic or inelastic collision? Fill in the momentum and kinetic energy values of block 1 and block 2 in the table below. Let movement to the left be negative. What was the net impulse acting on block 1 (with mass 4.220 kg)? How much work was done by the spring on block 2 (with mass 21.10 kg)? What was the maximum compression of the spring during the collision if the spring constant is k = 1600 N/m?Explanation / Answer
as there is no force is acting on the blocks, which can hold the two blocks together with spring compressed,
it is an elastic collision.
Q5)
before collision:
speed of m1= v1=+5 m/s
hence p1=m1*v1=21.1 kg.m/s
kinetic energy=0.5*m1*v1^2=52.75 J
speed of m2=v2=-2.5 m/s
hence p2=m2*v2=-52.75 kg.m/s
kinetic energy=KE2=0.5*m2*v2^2=65.9375 J
let after collision, speed of block 1 is v1 to the right and speed of block 2 is v2 to the right.
conserving momentum:
21.1-52.75=4.22*v1+21.1*v2
==>4.22*v1+21.1*v2=-31.65
==>v1=-5*v2-7.5....(1)
conserving total energy:
52.75+65.9375=0.5*4.22*v1^2+0.5*21.1*v2^2
==>2.11*v1^2+10.55*v2^2=118.6875
==>v1^2+5*v2^2=56.25
using v1=-5*v2-7.5
we get
25*v2^2+7.5^2+2*5*7.5*v2+5*v2^2=56.25
==>30*v2^2+75*v2+56.25=56.25
==>v2*(30*v2+75)=0
==>v2=0 or v2=-2.5 m/s
taking v2=-2.5 m/s,
we get v1=5 m/s
but this is not possible as it says that mass m1 will continue moving to the right and mass m2 will continue moving to the left.
hence v2=0
==>v1=-7.5 m/s
then block m1 will move to the left with 7.5 m/s and block m2 will come to rest.
hence after collision:
p1=-4.22*7.5=-31.65 kg.m/s
p2=0
Ke1=0.5*4.22*7.5^2=118.6875 J
JE2=0
6)net impulse=change in momentum=4.22*(-7.5-5)=-52.75 kg.m/s
7)work done by the spring on block 2=-(change in total energy of block 2)=-(0-65.9375)=65.9375 J
8)at maximum compression, speed of both the blocks will be 0 and initial kinetic energy of both the blocks will be converted to the potential energy of the spring
==>if maximum compression is d,
then 0.5*spring constant*d^2=118.6875
==>0.5*1600*d^2=118.6875
==>d=0.3851 m
hence maximum compression of the spring is 38.51 cm