Question

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

Answer 1

 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 = 2*m1/(m1 + m2)*v1 = 2*1.55/(1.55 + 4.70)*5.60 = 2.78m/s


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Answer 2

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

$h = \frac{v^2}{2g}$

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

$h_1 = 0.36 m$

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

$h_2 = 0.40 m$