Starting with an initial speed of 5.00 m/s at a height of 0.335 m, a 1.65-kg bal

Question

Starting with an initial speed of 5.00 m/s at a height of 0.335 m, a 1.65-kg ball swings downward and strikes a 4.45-kg ball that is at rest, as the drawing shows.
(a) Using the principle of conservation of mechanical energy, find the speed of the 1.65-kg ball just before impact.
1 m/s

(b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision.
2 m/s (1.65 kg-ball)
3 m/s (4.45 kg-ball)

(c) How high does each ball swing after the collision, ignoring air resistance?
4 m (1.65 kg-ball)
5 m (4.45 kg-ball)

Explanation / Answer

a.) Use the formula for conservation of mechanical energy: -1/2mv(1)^2+mgy(1)=1/2mv(2)^2+mgy(2) -the second potential energy term cancels because the height of the lighter ball is zero when it strikes the heavier ball. The mass also cancels. -1/2v(1)^2+gy(1)=1/2v(2)^2 solve for v(2) v(2)=sqrt(v(1)^2+gy(1)) sqrt(5^2 + 9.81(.335)) v(2)=5.318 m/s b.) Use conservation of momentum and the formula for head-on elastic collisions -m(1)v(1)+m(2)v(2)=m(1)v(1)'+m(2)v(2)' v(1)-v(2)=-(v(1)'-v(2)') use elimination or substitution to solve for v(1)' v(2)=0 because the second object is at rest. v(1)'=v(1)(m(1)-m(2))/(m(1)+m(2) -plug in numbers v(1)= 5.318 m/s v(1)'=(5.318)(1.65-4.45)/(1.65+4.45)) v(1)'=-2.441 m/s. This means that the first object bounces of the second and goes back the way it came. plug in this number into the formula for head-on elastic collisions v(1)-v(2)=-(v(1)'-v(2)') v(2)'=2.759 /ms c) use conservation of mechanical energy for each ball y(1)=0 each time. m cancels; solve for y(2). -y(2)=1/2(v(1)^2)/g light: -y(2)=1/2(-2.441^2)/9.81 heavy: -y(2)=1/2(2.759^2)/9.81 -Plug in numbers. The v(1) for each object are the ones calculated in part b. each v(2)=0 because they stop at the top of the swing. y(2) for the light mass =.304 m y(2) for the heavy mass =.388 m

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