Two blocks can collide in a one-dimensional collision. The block on the left has

Question

Two blocks can collide in a one-dimensional collision. The block on the left has a mass of 0.40 kg and is initially moving to the right at 2.4 m/s toward a second block of mass 0.80 kg that is initially at rest. When the blocks collide, a cocked spring releases 1.2 J of energy into the system. (For velocities, use + to mean to the right, - to mean to the left.) What is the velocity of the first block after the collision? What is the velocity of the second block after the collision? Remember that the blocks cannot pass through each other!

Explanation / Answer

Given

m1 = 0.40 kg, m2 = 0.80 kg, u1 = 2.4 m/s, u2 = 0 m/s

cocked energy released into the system is 1.2 J

from the conservation of momentum

m1u1 +m2u2 = m1v1 +m2v2 ==> m1u1 = m1v1 +m2v2,------>(A) u2 = 0 m/s

now due to 1.2 J of energy added in to the system the blocks move apart in the opposite directions

say m1 moves to the left and m2 moves to right directions

by conservation of energy

   0.5(m1u1^2)+1.2 = 0.5(m1v1^2)+0.5(m2v2^2)

   (m1u1^2)+2.4 = (m1v1^2)+(m2v2^2) ------------>(B)

from the eq (A), v2 = m1(u1-v1)/m2

  
now eq (B)===> (m1u1^2)+2.4 = (m1v1^2)+(m2v2^2)

       (m1u1^2)+2.4 = (m1v1^2)+(m2(m1((u1-v1)/m2)^2)

       (m1u1^2)+2.4 = m1v1^2 + (m1^2/m2)(u1-v1)^2

substituting the values solving for v1

       (0.4*2.4^2)+2.4 = 0.4*v1^2+(0.4^2/0.8)(2.4-v1)^2

       v1 = -1.76125 m/s

now v2 = 0.4(2.4+1.76125)/0.8 = 2.080 m/s

a) v1 = -1.76125 m/s

b) 2.080 m/s

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