Question

a 1.50 kg red ball traveling at 3.0m/s collides head on with a 3.50 kg green marble moving 1.5 m/s. What is the final velocity of each?

Answer 1

That's one massive marble



Like this? If we're assuming a completely elastic collision, then we know two things about the system. Momentum is conserved and kinetic energy is conserved. Two equations for two unknowns.

\[
m_{Ball} = m_1 = 1.5kg
\\v_{i \ Ball} = u_i = 3m/s
\\ v_{f \ Ball} = u_f = ?
\\ \
\\ m_{Marble}=m_2=3.5kg
\\ v_{i \ Marble} = v_i = -1.5 m/s
\\ v_{f \ Marble} = v_f = ?
\]
\[\underline{\textrm{Conservation of Momentum}}
\\ \
\\ \hspace{60px}p_i = p_f
\\ \
\\ m_1u_i + m_2v_i = m_1u_f + m_2v_f
\\ \
\\ m_2 v_f = \underbrace{(m_1u_i + m_2v_i)}_{\hbox{We know this value;}}_{\hbox{ call it "A"}} -m_1u_f
\\ \hspace{130px}A= (1.5kg)(3m/s)+(3.5kg)(-1.5m/s)
\\ \
\\m_2^2v_f^2 = (A-m_1u_f)^2
\\ \]


\[ \underline{\textrm{Conservation of Energy}}
\\ \
\\ \hspace{60px}K_i = K_f
\\ \
\\ \frac{1}{2}m_1u_i^2 + \frac{1}{2}m_2v_i^2 = \frac{1}{2}m_1u_f^2 + \frac{1}{2}m_2v_f^2
\\ \
\\ \
\\ m_1u_i^2 + m_2v_i^2 = m_1u_f^2 + m_2v_f^2
\\ \
\\ \ m_2v_f^2 = \underbrace{m_1u_i^2 + m_2v_i^2}_{\hbox{Call this "B"}} - m_1u_f^2
\\ \
\\ \ \hspace{110px} B = (1.5kg)(3.0m/s)^2+(3.5kg)(-1.5m/s)^2
\\ \
\\ \ m_2(m_2v_f^2)=m_2(B- m_1u_f^2)
\\ \
\\ \
\\ \
\\ \
\\ \ m_2(m_2v_f^2) = m_2^2v_f^2
\\ \
\\ m_2(B-m_1u_f^2) = (A-m_1u_f)^2
\]
Solve for u_f and then you have the key to finding v_f !

I only named A and B so that the final expression wasn't so gigantic ^_^