momentum problem help needed answer is in the picture (h_0*m_1)/(m_1+m_2) detail

Question

momentum problem help needed answer is in the picture (h_0*m_1)/(m_1+m_2)

detailed help please T__T

Thanks for your help in advance

971 A pendulum is fastened to a car on a horizontal frictionless surface, as shown. The pendulum ball, mass m2, is released from rest at height ho above its lowest position. The car, of mass mi, is held while the ball is descending, so the car does not move. At the instant the ball passes Awswer: through its lowest position, the car is released Find the height to which the ball rises on the other side of its swin (Hint: Note that when the ball is at its final high position, both ball and car are moving with the same velocity o d

Explanation / Answer

using energy conservation before car is released

KE1 + PE1 = KE2 + PE2

KE1 = 0

PE2 = 0

m2*g*h0 = 0.5*m2*v2^2

v2 = velocity at lowest position = sqrt(2*g*h0)

Now using momentum conservation when car is released

Pi = Pf

m2*v2 + m1*v1 = m2*v3 + m1v4

v1 = 0, as car is just released

m2*sqrt(2*g*h0) = m1*v4 + m2*v3

given v3 = v4 = v

m2*sqrt(2*g*h0) = (m1 + m2)*v

v = m2*sqrt(2*g*h0)/(m1 + m2)

using energy conservation after kar is released

KE of m1 + PE at lowest + KE of m2 = KE of m1 at final + PE at final + KE of m2 at final

0 + 0 + 0.5*m2*v2^2 = 0.5*m1*v^2 + m2gh + 0.5*m2*v^2

m2*g*h = 0.5*m2*(2*g*h0) - 0.5*v^2*(m1 + m2)

using v = m2*sqrt(2*g*h0)/(m1 + m2)

m2*g*h = m2*g*h0 - 0.5*(m1 + m2)[m2*sqrt(2*g*h0)/(m1 + m2)]^2

m2*g*h = m2*g*h0 - 0.5*(m1 + m2)*m2^2*g*h0/(m1 + m2)^2

m2*g*h = m2*g*h0 - m2^2*g*h0/(m1 + m2)

m2*g*h = [m2*(m1 + m2)*g*h0 - m2^2*g*ho]/(m1 + m2)

dividing by g

m2*h = [m1*m2*h0 + m2^2*h0 - m2^2*h0]/(m1+m2)

m2*h = m1*m2*h0/(m1+m2)

h = m1*h0/(m1+m2)

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