ment 02 /courses/1484261/quizzes/2007703/take Question 4 1 pts speed speed speed
ID: 1731909 • Letter: M
Question
ment 02 /courses/1484261/quizzes/2007703/take Question 4 1 pts speed speed speed speed Mass m 0.1 kg moves to the right with speed v-0.68 m/s and collides with an equal mass initially at rest. After this inelastic collision the system retains a fraction 0.81 of its original kinetic energy. If the masses remain in contact for 0.01 secs while colliding, what is the average force in N between the masses during the collision? Hints: All motion is in 1D. Ignore friction between the masses and the horizontal surface. You will probably need to use the quadratic formula to solve the resulting equations. VR must be greater than Vu since the masses can't pass through each other!Explanation / Answer
Initial velocity of the mass 1(v) =0.68 m/s
other mass is at rest therefore the initial velocity of other mass will be (u2) = 0
Therefore the initial momentum of the system = mv +mu2 = (0.1*0.68 ) +(0.1*0)
= 0.068 ---------(1)
Now the initial kinetic energy of the system
= (1/2)mv2 +(1/2)mu22 = (1/2)*0.1*(0.68)2 + 0 = 0.02312 J
Now since the collision is elastic therfore both block will stick after the collision ,
so let us consider their velocity wil be V
Hence the final momentum of the system = (m+m)V = 2mV -------------(2)
Now the final kinetic energy = (1/2)*(m+m)V2 = (1/2)*2m*V2 = mV2 ----------(3)
As we know that the final energy is = 0.81 times the original kinetic energy , therefore
mV2 = 0.81* 0.02312
0.1*V2 = 0.81* 0.02312
V = 0.433 m/s
Hence the final momentum will be = 2mV = 2*0.1*0.433 = 0.0866 kg-m/s
Now we know that
Force*time = Change in momentum
Force*time= (final momentum - Initial momentum)
Force*0.01 = 0.0866 - 0.068
Force = 1.86 N
hence the force will be 1.86 N
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