Ball 1, with a mass of 110 g and traveling at 12 m / s , collides head on with b

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Explanation / Answer

Let m1 and m2 be the masses of the two balls and u and 0 their initial velocities. Let v1 and v2 be their velocities after the collision in the direction of u.
Since the collision is head on, the motion of both balls is along a line.

By law of conservation of momentum,
m1*u + 0 = m1*v1 + m2*v2 ... ( 1 )

As the collision is elastic, kinetic energy is conserved
=> (1/2)m1*u^2 = (1/2)m1*v1^2 + (1/2)m2*v2^2
=> m1u^2 = m1v1^2 + m2v2^2 ... ( 2 )

Plugging v2 = (m1/m2)(u - v1) from eqn. ( 1 ) into eqn. ( 2 )
m1u^2 = m1v1^2 + m2*[(m1/m2)(u - v1)]^2
=> u^2 = v1^2 + (m1/m2)(u - v1)^2
=> u^2 - v1^2 = (m1/m2)(u - v1)^2
=> u + v1 = (m1/m2)(u - v1)
=> (u + v1)/(u - v1) = m1/m2
=> v1/u = (m1-m2)/(m1+m2)
=> v1 = u * (m1-m2)/(m1+m2)

u = 12, m1 = 110, m2 = 320

v1=5.86 m/sec is the fnal velocity of the ball n elastic condition

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