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Question 9.126: Particle 1 of mass 200 g and speed 3.00 m/s undergoes a one-......

Particle 1 of mass 200 g and speed 3.00 m/s undergoes a one-dimensional collision with stationary particle 2 of mass 400 g. What is the magnitude of the impulse on particle 1 if the collision is (a) elastic and (b) completely inelastic?

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Using Eq. 9-67, we have after the elastic collision

v_{1 f}=\frac{m_1-m_2}{m_1+m_2}v_{1 i}       (9-67)

v_{1 f}=\frac{m_1-m_2}{m_1+m_2}v_{1 i}=\frac{-200 \,g}{600\, g}v_{1 i}=-\frac{1}{3}(3.00\, m / s )=-1.00 \,m / s .

(a) The impulse is therefore

\begin{aligned}J &=m_1 v_{1 f}-m_1 v_{1 i}=(0.200 \,kg )(-1.00 \,m / s )-(0.200\, kg )(3.00 \,m / s )=-0.800 \,N \cdot s \\&=-0.800 \,kg m / s,\end{aligned}

or |J| = -0.800 kg.m/s.

(b) For the completely inelastic collision Eq. 9-75 applies

v_{1 f}=\frac{m_1-m_2}{m_1+m_2}v_{1 i}+\frac{2 m_2}{m_1+m_2}v_{2 i}            (9-75)

v_{1 f}=V=\frac{m_1}{m_1+m_2}v_{1 i}=+1.00 \,m / s.

Now the impulse is

\begin{aligned}J &=m_1 v_{1 f}-m_1 v_{1 i}=(0.200 \,kg )(1.00 \,m / s )-(0.200 \,kg )(3.00 \,m / s )=0.400 \,N \cdot s \\&=0.400 \,kg m / s\end{aligned}

.

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