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# Answer to Question #125743 in Mechanical Engineering for Kemah Ri

Question #125743
A 600 g ball A is moving with a velocity of magnitude 6 m/s when it hit by a 1 kg ball B that has a velocity of magnitude 4 m/s. Knowing that the coefficient of restitution is 0.8 and assuming no friction, determine the velocity of each ball after impact.
1
2020-07-22T06:03:36-0400

"(v_A)_n=6\\cdot \\cos40\u00b0=4.596m\/s"

"(v_A)_t=-6\\cdot \\sin40\u00b0=-3.857m\/s"

"(v_B)_n=-4m\/s"

"(v_B)_t=0"

t-direction:

"m_A(v_A)_t+m_B(v_B)_t=m_A(v'_A)_t+m_B(v'_B)_t"

"0.6\\cdot(-3.857)+0=0.6\\cdot(v'_A)_t+1\\cdot (v'_B)_t"

"-2.314=0.6\\cdot(v'_A)_t+(v'_B)_t"

For ball A

"m_A\\cdot(v_A)_t=m_A\\cdot(v'_A)_t\\to -3.857=(v'_A)_t\\to (v'_A)_t=-3.857m\/s"

So, we have

"-2.314=0.6\\cdot(-3.857)+(v'_B)_t\\to (v'_B)_t=0"

n-direction

"((v_A)_n-(v_B)_n)e=(v'_B)_n-(v'_A)_n"

"(4.596-(-4))\\cdot0.8=(v'_B)_n-(v'_A)_n"

"(v'_B)_n-(v'_A)_n=6.877\\to (v'_B)_n=6.877+(v'_A)_n"

"m_A(v_A)_n+m_B(v_B)_n=m_A(v'_A)_n+m_B(v'_B)_n"

"0.6\\cdot 4.596+1\\cdot(-4)=1\\cdot(v'_B)_n+0.6\\cdot(v'_A)_n"

"(v'_B)_n+0.6\\cdot(v'_A)_n=-1.2424\\to 6.877+(v'_A)_n+0.6\\cdot(v'_A)_n=-1.2424"

"(v'_A)_n=-5.075m\/s"

"(v'_B)_n=6.877+(v'_A)_n=6.877-5.075=1.802m\/s"

For ball A after impact

For ball B after impact

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