1- consider a 1-D collision between blocks of objects of m1 = 2 kg and m2 = 5 kg

Question

1- consider a 1-D collision between blocks of objects of m1 = 2 kg and m2 = 5 kg. the collision occurs along the x-direction. before the collision m1 has an x-component of velocity of 10 m/s and m2 has an x-component of velocity of -6 m/s. after the collision m1 has an x-component of velocity of -4 m/s.

A) Find the x-component of velocity of m2

B) find Q for this collision. recall Ktot,bef - Ktot,aft = Q

2- consider a 1-D collision between blocks of objects of m1 = 2 kg and m2 = 5 kg. the collision occurs along the x-direction. before the collision m1 has an x-component of velocity of 10 m/s and m2 has an x-component of velocity of -6 m/s.

A) If the objects stick together as a result of the collision, find the x-component of velocity of the two blocks after the collision.

B) if the objects stick together as a result of the collision, find the differece between the total kinetic energy of the blocks before the collision and the total kinetic energy of the blocks after the collision. that is find Q, where Ktot,bef = Ktot,aft + Q

3- consider a 1-D collision between blocks of objects of m1 = 2 kg and m2 = 5 kg. the collision occurs along the x-direction. before the collision m1 has an x-component of velocity of 10 m/s and m2 has an x-component of velocity of -6 m/s. find the x-component of the velocities of the two blocks after the collision if

A) Q = 0 (Recall Ktot,bef - Ktot,aft = Q)

B) Q = 100J

C) Q= 300J

4- find the coefficient of restitution for the collisions indicated in problem 3 parts A) and B)

Explanation / Answer

1)

m1 = 2 kg       m2 = 5 kg

u1x = 10 m/s     u2x = -6 m/s

after collision

v1x = -4 m/s       v2x = ?

initial momentum = final momentum

m1*u1x + m2*u2x = m1*v1x + m2*v2x

(2*10)-(5*6) = -(2*4)+(5*v2)

v2x = -0.4 m/s

+++++++++++

2)

m1 = 2 kg       m2 = 5 kg

u1x = 10 m/s     u2x = -6 m/s

after collision

v1x = v2x =v ?

initial momentum = final momentum

m1*u1x + m2*u2x = (m1+m2)*v

(2*10)-(5*6) = (2+5)*v

v = -1.43 m/s

(B)

before Ki = 0.5*m1*u1x^2 + 0.5*m2*u2x^2

Ki = (0.5*2*10^2)+(0.5*5*6^2) = 190 J

Kf = 0.5*m1*v1x^2 + 0.5*m2*v2x^2

Kf = (0.5*2*1.43^2)+(0.5*5*1.43^2) = 7.16 J

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