1. Determine the velocity of the m 2 block just after the collision. (Use positi

Question

1. Determine the velocity of the m2 block just after the collision. (Use positive sign if the motion is to the right, negative if it is to the left.)

2. Determine the velocity of the m1 block just after the collision. (Use positive sign if the motion is to the right, negative if it is to the left.)

3. Determine the maximum height to which m2 rises after the collision.

4. Determine the maximum height to which m2 rises after the collision.

Two blocks of masses m1 = 1.91 kg and m2 = 3.82 kg are each released from rest at a height of h = 4.89 m on a frictionless track, as shown in the figure, and undergo an elastic head-on collision. Im m2

Explanation / Answer

Before collision teh speed of m1 is u1

and that of m2 is u2

then

apply law o conservation of energy

0.5*m1*u1^2 = m1*g*h

u1 = sqrt(2*g*h) = sqrt(2*9.8*4.89) = 9.8 m/s

similarly u2 = 9.8 m/s

let v1 and v2 be the speeds of the m1 and m2 after collision respectievly

v1 = (m1-m2)*u1/(m1+m2) -(2*m2*u2)/(m1+m2)

v1 = (m1-2m1)*u1/(3*m1) - (2*2*m1*u1)/(3*m1)

v1 = (-u1/3)-(4*u1/3) = -5*9.8/3 = -16.33 m/s

V2 = (m2-m1)*u1/(m1+m2) -(2*m1*u1)/(m1+m2)

v2 = (u1/3)-(2*u1/3) = -u1/3 = -9.8/3 = -3.26 m/s

agian from law of conservation of energy

0.5*m2*v2^2 = m2*g*h2

h1 = 0.5*v2^2/g = 0.5*3.26^2/(9.8) = 0.54 m

-----------------------------------

h1 = 0.5*v1^2/g = 0.5*16.33^2/9.8 = 13.6 m

answers are

1) v2 =-3.26 m/s

2) v1 = -16.33 m/s

3) h2 = 0.54 m

4) h1 = 13.6 ...this is for m1

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