Question

Derive equation Vb=M/m*sqrt2gRcm(1-cos)

Answer 1

when a ball of mass m is pushed by muzzle witha velocity v0 at a pendulum bob of mass M

which is suspended by a rod of length R and negligible mass. After the collision, the pendulum and ball stick together and swing to a maximum angular displacement θ as shown.

                                                Since momentum is conserved, we can use that as a starting point:

                p0 = mv0

                pf = (m+M)v

                Since momentum is conserved, we can say that:

                p0 = pf

Therefore:

                mv0 = (m+M)v

                and:

                v0 = ((m+M)v / m)

here energy conservation when the bod and the ball are moving to a certain height

                                U = (m+M)gh

                                KE = 1/2(m+M)v2

                                (m+M)gh = 1/2mv2

                                v = sqrt(2gh),--------- A    we can find h in terms of the angle θ and the length of the pendulum, R:

                                h = R(1 – cos(θ))

putting the values of h in eq A

                                v = sqrt(2gR(1-cos(θ)))

now substituting the v in v0 we get

                                v0 = ((m+M)v / m)

                           v0 = ((m+M)/m)(v)

                            v0 = ((m+M)/m)(sqrt(2gR(1-cos(θ))))

here v0 is velocity of the ball vb and r is length of the pendulum Rcm

       

                                vb = ((m+M)/m)(sqrt(2gRcm(1-cos(θ))))