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Help!!!! Please..... Someone. Problem:


Starting with an initial speed of 5.00 m/s at a height of h = 0.260 m, the m1 = 1.55 kg ball swings downward and strikes the m2 = 4.50 kg ball that is at rest, as the drawing shows.

AUsing the principle of conservation of mechanical energy, find the speed of the 1.55-kg ball just before impact.
BAssuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision.
CHow high does each ball swing after the collision, ignoring air resistance? ...?

Verified Answer :

For an elastic collision with a stationary object we have the following 


v1a = (m1 - m2)/(m1 + m2) v1 where v1 is the velocity of the incoming ball and v1a is its velocity after the collsion 


so for ball1 we have v1a = (1.55 - 4.70)/(1.55 + 4.70)5.60m/s = -2.82 m/s (The negative indicates ball 1 is moving in the opposite direction) 


For ball 2 = have v2 = 2m1/(m1 + m2)v1 = 21.55/(1.55 + 4.70)5.60 = 2.78m/s


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Answer:

Part a)

\(v_f = 5.48 m/s\)

Part b)

\(v_1 = -2.67 m/s\)

\(v_2 = 2.81 m/s\)

Part c)

\(h_1 = 0.36 m\)

\(h_2 = 0.40 m\)

Explanation:

Part a)

As we know by energy conservation law

initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

So here we know that

\(\frac{1}{2}mv_i^2 + mgh_i = \frac{1}{2}mv_f^2 + mgh_f\)

\(v_i = 5 m/s\)

\(h_i = 0.260 m\)

\(h_f = 0\)

now we have

\(v_f^2 = v_i^2 + 2gh\)

\(v_f^2 = 5^2 + 2(9.8)(0.260)\)

\(v_f = 5.48 m/s\)

Part b)

As we know that collision is perfectly elastic collision

so we will have

\(m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}\)

\(1.55(5.48) + 0 = 1.55 v_1 + 4.50 v_2\)

also we know for elastic collision

\(v_2 - v_1 = 5.48\)

\(v_1 = -2.67 m/s\)

\(v_2 = 2.81 m/s\)

Part c)

Now the height of each ball is given by

\(h = \frac{v^2}{2g}\)

\(h_1 = \frac{2.67^2}{2(9.81)}\)

\(h_1 = 0.36 m\)

\(h_2 = \frac{2.81^2}{2(9.81)}\)

\(h_2 = 0.40 m\)

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