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 4.55-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.
m/s

(b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision.
m/s (1.65 kg-ball)
m/s (4.55 kg-ball)

(c) How high does each ball swing after the collision, ignoring air resistance?
m (1.65 kg-ball)
m (4.55 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)
-plug in the numbers

v(1) =5.31 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)= 5.31m/s
v(1)'=-2.48 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 and get
v(2)'

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
-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.

=1.437

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