Answer to Question #125743 in Mechanical Engineering for Kemah Ri

Question

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.

Answer 1

$(v_A)_n=6\cdot \cos40°=4.596m/s$

$(v_A)_t=-6\cdot \sin40°=-3.857m/s$

$(v_B)_n=-4m/s$

$(v_B)_t=0$

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

$\tan\beta=\frac{(v_A)_t}{(v_A)_n}=\frac{3.857}{5.075}\to \beta=37.2°\to \beta+40°=77.2°$ Answer

$v'_A=\sqrt{3.857^2+5.075^2}=6.37m/s$ Answer

For ball B after impact

$v'_B=1.802m/s$, $\gamma=40°$ Answer