A ball of mass m1 moves at speed vi along the positive x-axis toward a second ba

Question

A ball of mass m1 moves at speed vi along the positive x-axis toward a second ball of mass m2, which is initially at rest. The second ball has 7 times the mass of the first ball. After the collision between these two objects, m1 moves along the positive y-axis at a speed v1, and m2 moves at a speed v2 = (1/6)vi at an angle of 31.0° below the positive x-axis.
(a) Find the momentum change of the ball of mass m1 during the collision. Give your answer in x- and y-component form; express the components in terms of m1 and v1.
?p1x
?p1y


(b) Repeat for the ball of mass m2.
?p2x
?p2y

Explanation / Answer

conserving momentum,
m1*vi(i) + 0 = m1*v1(j) + 7*m1*vi/6(cos31(i) - sin31(j))

equating the x and y components,we get

0=m1*v1 - (7/6)*m1*vi*cos31
so, v1 = (7/6)*cos31*vi = vi

so, change in momentum of mass m1 =m1*v1(j) - m1*vi(i) = -m1*vi(i) +m1*vi(j)

so, p1x = -m1*vi  

     p1y = m1*vi

 

 

so, change in moemntum of mass m2 = 7*m1*(vi/6)(cos31(i) - sin31(j))

so ,

p2x = m1*vi  

 p2y = -0.6*m1*vi

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