Home Page Textbooks Physics: Principles and Problems Page 238: Practice Problems

Physics: Principles and Problems

9th Edition

Elliott, Haase, Harper, Herzog, Margaret Zorn, Nelson, Schuler, Zitzewitz

ISBN: 9780078458132

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Page 238: Practice Problems

Exercise 13

Step 1

1 of 2

Given: $m=3cdot10^{5},,rm kg$, $v_1=2.2,,rm m/s$

We can use conservation of momentum:

$$
begin{align*}
mv_1&=2mv\
v&=frac{v_1}{2}\
v&=frac{2.2}{2}
end{align*}
$$

$$
boxed{v=1.1,,rm m/s}
$$

Result

2 of 2

$$
v=1.1,,rm m/s
$$

Exercise 14

Step 1

1 of 2

Given: $m=0.105,,rm kg$, $v_1=24,,rm m/s$, $M=75,,rm kg$

We can use conservation of momentum:

$$
begin{align*}
mv_1&=(M+m)v\
v&=v_1frac{m}{M+m}\
&=24cdotfrac{0.105}{75+0.105}
end{align*}
$$

$$
boxed{v=0.034,,rm m/s}
$$

Result

2 of 2

$$
v=0.034,,rm m/s
$$

Exercise 15

Step 1

1 of 2

Given: $m=35,,rm g$, $M=5,,rm kg$, $v=8.6,,rm m/s$

We can use conservation of momentum:

$$
begin{align*}
mv_1&=(M+m)v\
v_1&=vfrac{M+m}{m}\
&=8.6cdotfrac{5+0.035}{0.035}
end{align*}
$$

$$
boxed{v_1=1237,,rm m/s}
$$

Result

2 of 2

$$
v_1=1237,,rm m/s
$$

Exercise 16

Step 1

1 of 2

Given: $m=35,,rm g$, $v_1=475,,rm m/s$, $M=2.5,,rm kg$, $v_1’=275,,rm m/s$

WE can use conservation of momentum:

$$
begin{align*}
mv_1&=Mv_2’+mv_1’\
v_2’&=frac{m(v_1+v_1′)}{M}\
&=frac{0.035cdot(475-275)}{2.5}
end{align*}
$$

$$
boxed{v_2’=2.8,,rm m/s}
$$

Result

2 of 2

$$
v_2’=2.8,,rm m/s
$$

Exercise 17

Solution 1

Solution 2

Step 1

1 of 2

Result

2 of 2

6.7 m/s

Step 1

1 of 5

**Given:**
– 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;

Step 2

2 of 5

Linear momentum is a measure of how difficult it is to change the state of motion of an object. The magnitude of the linear momentum is given as:
$$p = m v$$

Step 3

3 of 5

We assume that this is a perfect elastic collision, where objects do not stick together. In this case, each object moves in a different direction, and come to rest some time after colliding. This way the objects are conserving momentum. We are interested in the velocity $v_2$ of the ball. We can write the law of momentum conservation as:
$$begin{align*}
m_1 v_{1, 0} + m_2 v_{2, 0} &= m_1 v_1 + m_2 v_2\
m_2 v_2 &= m_1 v_{1, 0} + m_2 v_{2, 0} – m_1 v_1
end{align*}$$

Step 4

4 of 5

Solving the equation above for $v$ we divide it by $m_2$. Take into account that the bullet bounces backward after the collision, and that its returning velocity $v_1$ is negative. Also, the initial velocity of the ball is zero so the term $m_2v_{2, 0}$ vanishes:
$$begin{align*}
v_2 &= frac{ m_1 (v_{1, 0} – v_1)}{ m_2} \
&= dfrac{35 mathrm{~g} cdot left( 475 ,frac{text{m}}{text{s}} – left( – 5 ,frac{text{m}}{text{s}} right) right) }{ 2.5 mathrm{~kg}} \
&= dfrac{0.035 mathrm{~kg} cdot left( 475 ,frac{text{m}}{text{s}} – left( – 5 ,frac{text{m}}{text{s}} right) right) }{ 2.5 mathrm{~kg}} \
&= 6.72 ,frac{text{m}}{text{s}}
end{align*}$$

$$boxed{ v_2 = 6.72 ,frac{text{m}}{text{s}} }$$

Result

5 of 5

$$v_2 = 6.72 ,frac{text{m}}{text{s}}$$

Exercise 18

Step 1

1 of 2

Given: $m_1=0.5,,rm kg$, $v_1=6,,rm m/s$, $m_2=1,,rm kg$, $v_2=12,,rm m/s$, $v_1’=14,,rm m/s$

We can use conservation of momentum:

$$
begin{align*}
m_1v_1+m_2v_2&=m_1v_1’+m_2v_2’\
v_2’&=frac{m_1(v_1-v_1′)+m_2v_2}{m_2}\
&=frac{0.5cdot(6-14)+1cdot12}{1}
end{align*}
$$

$$
boxed{v_2’=8,,rm m/s}
$$

Result

2 of 2

$$
v_2’=8,,rm m/s
$$

Exercise 19

Step 1

1 of 2

Given: $M=4,,rm kg$, $m=50,,rm g$, $v_1=625,,rm m/s$

We can use conservation of momentum:

$$
begin{align*}
(M-m)v&=mv_1\
v&=v_1frac{m}{M-m}\
&=625frac{0.05}{4-0.05}
end{align*}
$$

$$
boxed{v=7.9,,rm m/s}
$$

Result

2 of 2

$$
v=7.9,,rm m/s
$$

Exercise 20

Step 1

1 of 2

Given: $m=1.5,,rm kg$, $M=4.5,,rm kg$, $v_1=27,,rm cm/s$

We can use conservation of momentum:

$$
begin{align*}
mv_1&=Mv_2\
v_2&=v_1frac{m_1}{m_2}\
&=27cdotfrac{1.5}{4.5}
end{align*}
$$

$$
boxed{v_2=9,,rm cm/s}
$$

Result

2 of 2

$$
v_2=9,,rm cm/s
$$

Exercise 21

Step 1

1 of 2

Given: $m=80,,rm kg$, $v_1=4,,rm m/s$, $M=115,,rm kg$

We can use conservation of momentum:

$$
begin{align*}
mv_1&=Mv_2\
v_2&=v_1frac{m}{M}\
&=4cdotfrac{80}{115}
end{align*}
$$

$$
boxed{v_2=2.8,,rm m/s}
$$

Result

2 of 2

$$
v_2=2.8,,rm m/s
$$

Exercise 22

Step 1

1 of 2

Given: $m_1=925,,rm kg$, $v_1=20.1,,rm m/s$, $m_2=1865,,rm kg$, $v_2=13.4,,rm m/s$

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.

Result

2 of 2

$v=11.2,,rm m/s$

$phi=53^o$ from north to west

Exercise 23

Step 1

1 of 2

Given: $m_1=1383,,rm kg$, $v_1=11.2,,rm m/s$, $m_2=1732,,rm kg$, $v_2=31.3,,rm m/s$

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.

Result

2 of 2

$v=18.1,,rm m/s$

$phi=74^o$ from south to east

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Page 238: Practice Problems

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