In symbols, linear momentum is expressed as p = mv.
PHYSICS PROBLEM SOLVING - LINEAR MOMENTUM
Question 1. A helium atom, mass 4u travels with non relativistic speed v normal to the surface of a certain material, makes an elastic collision with an (essentially free) surface atom, and leaves in the opposite direction with speed 0.6v. The atom on the surface must be an atom of A. Hydrogen, mass 1u
B. Helium, mass 4u
C. Carbon, mass 12u
D. Oxygen, mass 16u
E. Silicon, mass 28u
Physics Solution: A
ma = 4u
va = v
vb = 0
va' = − 0.6v
Conservation of momentum of the system:
\[\begin{align*} m_av_a + m_bv_b &= m_av_a' + m_bv_b' \\4uv+0 &= 4u(-0.6v) + m_bv_b'\\ 4uv &= -2.4uv + m_bv_b'\\m_bv_b' &= 6.4uv....(1) \end{align*}\]
restitution coefficient:
\[\begin{align*} e&=-\left ( \frac{v'_a-v'_b}{v_a-v_b}\right )\\1&=-\left ( \frac{-0.6v-v'_b}{v-0}\right )\\v&=0.6v+v'_b\\v'_b&=0.4v....(2) \end{align*}\](2) to (1) substitution:\[\begin{align*} m_bv_b' &= 6.4uv\\m_b(0.4v) &= 6.4uv\\m_b&=\frac{6.4}{0.4}u=16u \end{align*}\]
Question 2. A man of mass m on an initially stationary boat gets off the boat by leaping to the left in an exactly horizontal direction. Immediately after the leap, the boat of mass M is observed to be moving to the right at speed v. How much work did the man do during the leap (both on his own body and on the boat)?
A. $\frac {1}{2}mv^2$
B. $\frac {1}{2}Mv^2$
C. $\frac {1}{2}(M+m)v^2$
D. $\frac{1}{2}\left (\frac{Mm}{M+m} \right )v^2$
E. $\frac{1}{2}\left (\frac{M^2}{m}+M \right )v^2$
Physics Solution: E\[\begin{align*} mv_o&=Mv\\v_o&=\frac{M}{m}v\\\\W&=\frac{1}{2}mv_o^2+\frac{1}{2}Mv^2\\W&=\frac{1}{2}m\left (\frac{M}{m}v \right )^2+\frac{1}{2}Mv^2\\W&=\frac{1}{2}\left (\frac{M^2}{m}+M \right )v^2\end{align*}\]
B. $\frac {1}{2}Mv^2$
C. $\frac {1}{2}(M+m)v^2$
D. $\frac{1}{2}\left (\frac{Mm}{M+m} \right )v^2$
E. $\frac{1}{2}\left (\frac{M^2}{m}+M \right )v^2$
Physics Solution: E\[\begin{align*} mv_o&=Mv\\v_o&=\frac{M}{m}v\\\\W&=\frac{1}{2}mv_o^2+\frac{1}{2}Mv^2\\W&=\frac{1}{2}m\left (\frac{M}{m}v \right )^2+\frac{1}{2}Mv^2\\W&=\frac{1}{2}\left (\frac{M^2}{m}+M \right )v^2\end{align*}\]
