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A girl riding a bicycle along a straight road with a speed of $5 \mathrm{~ms}^{-1}$ throws a stone of mass $0.5 \mathrm{~kg}$ which has a speed of $15 \mathrm{~ms}^{-1}$ with respect to the ground along her direction of motion. The mass of the girl and bicylce is $50 \mathrm{~kg}$. Does the speed of the bicycle change after the stone is thrown? What is the change in speed, if so?
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Ag given that,
Velocity of bicycle, $u_1=5 \mathrm{~m} / \mathrm{s}$,
Velocity of stone $u_2=15 \mathrm{~m} / \mathrm{s}$,
Mass of stone $m_2=0.5 \mathrm{~kg}$
Mass of girl and bicycle $m_1=50 \mathrm{~kg}$
Total mass of girl, bicycle and stone
$(\mathrm{m})=(50+0.5) \mathrm{kg}=50.5 \mathrm{~kg}$.
Let after throwing the stone the speed of bicycle will change by $v_1 \mathrm{~m} / \mathrm{s}$.
According to law of conservation of linear momentum, Initial momentum (Girl, cycle, body)
$=$ Final momentum (cycle $+$ girl) and body
$m u_1=m_2 u_2+m_1 v_1$
$\left(m_1+m_2\right) u_1=m_2 u_2+m_1 v_1$
$50.5 \times 5=0.5 \times 15+50 \times v_1$
$252.5-7.5=50 v_1$
or $v_1=\frac{245.0}{50}=4.9 \mathrm{~m} / \mathrm{s}$
Change in speed $=5-4.9=0.1 \mathrm{~m} / \mathrm{s}$.
Velocity of bicycle, $u_1=5 \mathrm{~m} / \mathrm{s}$,
Velocity of stone $u_2=15 \mathrm{~m} / \mathrm{s}$,
Mass of stone $m_2=0.5 \mathrm{~kg}$
Mass of girl and bicycle $m_1=50 \mathrm{~kg}$
Total mass of girl, bicycle and stone
$(\mathrm{m})=(50+0.5) \mathrm{kg}=50.5 \mathrm{~kg}$.
Let after throwing the stone the speed of bicycle will change by $v_1 \mathrm{~m} / \mathrm{s}$.
According to law of conservation of linear momentum, Initial momentum (Girl, cycle, body)
$=$ Final momentum (cycle $+$ girl) and body
$m u_1=m_2 u_2+m_1 v_1$
$\left(m_1+m_2\right) u_1=m_2 u_2+m_1 v_1$
$50.5 \times 5=0.5 \times 15+50 \times v_1$
$252.5-7.5=50 v_1$
or $v_1=\frac{245.0}{50}=4.9 \mathrm{~m} / \mathrm{s}$
Change in speed $=5-4.9=0.1 \mathrm{~m} / \mathrm{s}$.
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