All Solutions
Page 238: Practice Problems
We can use conservation of momentum:
$$
begin{align*}
mv_1&=2mv\
v&=frac{v_1}{2}\
v&=frac{2.2}{2}
end{align*}
$$
$$
boxed{v=1.1,,rm m/s}
$$
v=1.1,,rm m/s
$$
We can use conservation of momentum:
$$
begin{align*}
mv_1&=(M+m)v\
v&=v_1frac{m}{M+m}\
&=24cdotfrac{0.105}{75+0.105}
end{align*}
$$
$$
boxed{v=0.034,,rm m/s}
$$
v=0.034,,rm m/s
$$
We can use conservation of momentum:
$$
begin{align*}
mv_1&=(M+m)v\
v_1&=vfrac{M+m}{m}\
&=8.6cdotfrac{5+0.035}{0.035}
end{align*}
$$
$$
boxed{v_1=1237,,rm m/s}
$$
v_1=1237,,rm m/s
$$
WE can use conservation of momentum:
$$
begin{align*}
mv_1&=Mv_2’+mv_1’\
v_2’&=frac{m(v_1+v_1′)}{M}\
&=frac{0.035cdot(475-275)}{2.5}
end{align*}
$$
$$
boxed{v_2’=2.8,,rm m/s}
$$
v_2’=2.8,,rm m/s
$$
– Mass: $m_1 = 35 mathrm{~g}$;
– Velocity: $v_{1, 0} = 475 ,frac{text{m}}{text{s}}$;
– Mass: $m_2 = 2.5 mathrm{~kg}$;
– Velocity: $v_{2, 0} = 0$;
– Velocity: $v_1 = 5 ,frac{text{m}}{text{s}}$;
**Required:**
– The velocity $v_2$ of the ball after the collision;
$$p = m v$$
$$begin{align*}
m_1 v_{1, 0} + m_2 v_{2, 0} &= m_1 v_1 + m_2 v_2\
m_2 v_2 &= m_1 v_{1, 0} + m_2 v_{2, 0} – m_1 v_1
end{align*}$$
$$begin{align*}
v_2 &= frac{ m_1 (v_{1, 0} – v_1)}{ m_2} \
&= dfrac{35 mathrm{~g} cdot left( 475 ,frac{text{m}}{text{s}} – left( – 5 ,frac{text{m}}{text{s}} right) right) }{ 2.5 mathrm{~kg}} \
&= dfrac{0.035 mathrm{~kg} cdot left( 475 ,frac{text{m}}{text{s}} – left( – 5 ,frac{text{m}}{text{s}} right) right) }{ 2.5 mathrm{~kg}} \
&= 6.72 ,frac{text{m}}{text{s}}
end{align*}$$
$$boxed{ v_2 = 6.72 ,frac{text{m}}{text{s}} }$$
We can use conservation of momentum:
$$
begin{align*}
m_1v_1+m_2v_2&=m_1v_1’+m_2v_2’\
v_2’&=frac{m_1(v_1-v_1′)+m_2v_2}{m_2}\
&=frac{0.5cdot(6-14)+1cdot12}{1}
end{align*}
$$
$$
boxed{v_2’=8,,rm m/s}
$$
v_2’=8,,rm m/s
$$
We can use conservation of momentum:
$$
begin{align*}
(M-m)v&=mv_1\
v&=v_1frac{m}{M-m}\
&=625frac{0.05}{4-0.05}
end{align*}
$$
$$
boxed{v=7.9,,rm m/s}
$$
v=7.9,,rm m/s
$$
We can use conservation of momentum:
$$
begin{align*}
mv_1&=Mv_2\
v_2&=v_1frac{m_1}{m_2}\
&=27cdotfrac{1.5}{4.5}
end{align*}
$$
$$
boxed{v_2=9,,rm cm/s}
$$
v_2=9,,rm cm/s
$$
We can use conservation of momentum:
$$
begin{align*}
mv_1&=Mv_2\
v_2&=v_1frac{m}{M}\
&=4cdotfrac{80}{115}
end{align*}
$$
$$
boxed{v_2=2.8,,rm m/s}
$$
v_2=2.8,,rm m/s
$$
We can use conservation of momentum in south north direction:
$$
begin{align*}
m_1v_1&=(m_1+m_2)v_y\
v_y&=frac{m_1v_1}{m_1+m_2}\
v_y&=frac{925cdot20.1}{925+1865}\
v_y&=6.7,,rm m/s
end{align*}
$$
We can use conservation of momentum in east west direction:
$$
begin{align*}
m_2v_2&=(m_1+m_2)v_x\
v_x&=frac{m_2v_2}{m_1+m_2}\
v_x&=frac{1865cdot13.4}{925+1865}\
v_x&=8.96,,rm m/s
end{align*}
$$
Now we can get speed:
$$
v=sqrt{v_x^2+v_y^2}=sqrt{6.7^2+8.96^2}
$$
$$
boxed{v=11.2,,rm m/s}
$$
And direction is:
$$
tanphi=frac{v_x}{v_y}=frac{8.96}{6.7}
$$
$$
boxed{phi=53^o}
$$
They move in direction of $53^o$ from north to west.
$phi=53^o$ from north to west
We can use conservation of momentum in north south direction:
$$
begin{align*}
m_1v_1&=(m_1+m_2)v_y\
v_y&=frac{m_1v_1}{m_1+m_2}\
v_y&=frac{1383cdot11.2}{1383+1732}\
v_y&=4.97,,rm m/s
end{align*}
$$
We can use conservation of momentum in west east direction:
$$
begin{align*}
m_2v_2&=(m_1+m_2)v_x\
v_x&=frac{m_2v_2}{m_1+m_2}\
v_x&=frac{1732cdot31.3}{1383+1732}\
v_x&=17.4,,rm m/s
end{align*}
$$
Now we can get speed:
$$
v=sqrt{v_x^2+v_y^2}=sqrt{4.97^2+17.4^2}
$$
$$
boxed{v=18.1,,rm m/s}
$$
And direction is:
$$
tanphi=frac{v_x}{v_y}=frac{17.4}{4.97}
$$
$$
boxed{phi=74^o}
$$
They move in direction of $4.97^o$ from south to east.
$phi=74^o$ from south to east
