Two blocks of masses m 1 = 1.96 kg and m 2 = 4.66 kg are each released from rest

Question

Two blocks of masses m1 = 1.96 kg and m2 = 4.66 kg are each released from rest at a height of h = 5.20 m on a frictionless track, as shown in the figure, and undergo an elastic head-on collision.

A) 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.)

B) 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.)

Explanation / Answer

here,

m1 = 1.96 kg

m2 = 4.66 kg

h = 5.2 m

let the velocity before the impact be u

using conservation of energy

0.5 m * v^2 = m * g * h

u^2 = 2 * 9.8 * 5.2

u = 10.09 m/s

u1 = u = 10.09 m/s

u2 = - u = - 10.09 m/s

let the velocity after collison be v1 amd v2

using conservation of momentum

m1*v1 +m2*v2 = m1*u1 + m2*u2

1.96 * v1 + 4.66 * v2 = 10.09 (1.96 - 4.66) ....... (1)

using conservation of kinetic energy

0.5 ( m1 * v1^2 + m2 * v2^2) = 0.5 (m1 * u1^2 + m2 * u2^2)

0.5 ( 1.96 * v1^2 + 4.66 * v2^2) = 0.5 (1.96 * (10.09)^2 + 4.66 * ( - 10.09)^2) .........(2)

solving both the equation for v1 and v2

v1 = - 18.32 m/s

v2 = 1.85 m/s

(A)
the velocity of the m2 block just after the collision is 1.85 m/s

(B)

the velocity of the m1 block just after the collision is - 18.32 m/s

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