Starting with an initial speed of 5.22 m/s at a height of 0.394 m, a 2.94 kg bal

Question

Starting with an initial speed of 5.22 m/s at a height of 0.394 m, a 2.94 kg ball swings downward and strikes a 5.63 kg ball that is at rest. a) using the principle of conservation of mechanical energy, find the speed of the 2.94 kg ball just before impact. b) assuming that the collision is elastic, find the velocity (magnitude and direction) of the 2.94 kg ball just after the collision. c) assuming that the collision is elastic, find the velocity (magnitude and direction) of the 5.63 kg ball just after the collision. d) how high does the 2.94 ball swing after the collision, ignoring air resistance? e) how high does the 5.63 kg ball swing after the collision, ignoring air resistance?

Explanation / Answer

a. E mech initial = E mech final Ugrav + KE = Ugrav + KE m*g*h +.5mv^2 = m*g*h + .5mv^2 2.94*9.8*.394 + 0 = 0 + .5*.394*v^2 V= 7.59 m/s b. for and elastic collision with a stationary object we use v1a = (m1 - m2)/(m1 + m2) *v1 where v1 is the velocity of the incoming ball and v1a is its velocity after the collision v1a=(2.94-5.63)/(2.94+5.63) 7.59 v1a= -2.63 ( the negative means its moving in the opposite direction. c. v2 = 2*m1/(m1 + m2)*v1 v2 = (2*2.94)/(2.94 + 5.63) 7.59 v2= 5.207 d. E mech initial = E mech final Ugrav + KE = Ugrav + KE m*g*h +.5mv^2 = m*g*h + .5mv^2 E 0+ .5(2.94)(-2.63^2) = (2.94)(9.8)(h) + 0 h= .3529m e. E mech initial = E mech final Ugrav + KE = Ugrav + KE m*g*h +.5mv^2 = m*g*h + .5mv^2 E 0 + .5(5.63)(5.207^2) = (5.63)(9.8)(h)+ 0 h = 1.383m If i made a mistake in my math those are the equations you use for solving this.

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