AP Physics C: Mechanics practice test 7.

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Question 1 |

### Find the force of gravity between two-point particles of mass $2.5 \ kg$ and $3.0 \ kg$ which are at a distance of $10 \ cm$ from each other.

$50.03×10^{-9} \ N$ | |

$50.03×10^{-11} \ N$ | |

$0.50×10^{-8} \ N$ | |

$5.03×10^{-9} \ N$ | |

$0.05×10^{-10} \ N$ |

Question 1 Explanation:

$F_g=\dfrac{Gm_1 m_2}{r^2}$

Using $G=6.67×10^{-11}, \ m_1=2.5 \ kg,m_2=3.0 \ kg$ and $r=0.1 \ m$ in the above equation gives,

$F_g=\dfrac{6.67×10^{-11}×2.5×3}{0.1^2} = 50.03×10^{-9} \ N$

Using $G=6.67×10^{-11}, \ m_1=2.5 \ kg,m_2=3.0 \ kg$ and $r=0.1 \ m$ in the above equation gives,

$F_g=\dfrac{6.67×10^{-11}×2.5×3}{0.1^2} = 50.03×10^{-9} \ N$

Question 2 |

### Three stars in the Milky Way were found in the following symmetrical arrangement.

### What is the net force acting on star J due to the other two stars?

### (The mass of each star is equal to $6 × 10^{28} \ kg$ and the $XY$ axis is in standard orientation)

$(9.68 \skew{2.3}\hat{i} - 29.20 \skew{4.5}\hat{j} )×10^{24} \ N$ | |

$(9.60 \skew{2.3}\hat{i} + 29.25 \skew{4.5}\hat{j} )×10^{25} \ N$ | |

$(29.20\skew{2.3}\hat{i} + 9.60 \skew{4.5}\hat{j} )×10^{27} \ N$ | |

$(6.78 \skew{2.3}\hat{i} - 26.38 \skew{4.5}\hat{j} )×10^{27} \ N$ | |

$(19.45 \skew{2.3}\hat{i} - 9.20 \skew{4.5}\hat{j} )×10^{26} \ N$ |

Question 2 Explanation:

Force on $J$ due to $K =\dfrac{Gm^2}{r^2}$

$=\dfrac{6.67×10^{-11}×36×10^{56}}{12.25×10^{18} }$

$=19.6×10^{27} \ N$ (along $-\skew{5}\hat{j}$)

Force on $J$ due to $L=\dfrac{Gm^2}{r^2}$

$=\dfrac{6.67×10^{-11}×36×10^{56}}{25×10^{18}}$

$=9.60×10^{27} \ N$ (along $\left(\dfrac{\skew{2}\hat{i}}{\sqrt{2}}-\dfrac{\skew{5}\hat{j}}{\sqrt{2}}\right)$)

Net force $=(6.78\skew{2}\hat{i}-26.38\skew{5}\hat{j} )×10^{27} \ N$

$=\dfrac{6.67×10^{-11}×36×10^{56}}{12.25×10^{18} }$

$=19.6×10^{27} \ N$ (along $-\skew{5}\hat{j}$)

Force on $J$ due to $L=\dfrac{Gm^2}{r^2}$

$=\dfrac{6.67×10^{-11}×36×10^{56}}{25×10^{18}}$

$=9.60×10^{27} \ N$ (along $\left(\dfrac{\skew{2}\hat{i}}{\sqrt{2}}-\dfrac{\skew{5}\hat{j}}{\sqrt{2}}\right)$)

Net force $=(6.78\skew{2}\hat{i}-26.38\skew{5}\hat{j} )×10^{27} \ N$

Question 3 |

### A satellite is moving in a circular orbit at a fixed distance of $1,000 \ km$ from the surface of the Earth. What is the time period of this satellite?

### $(M_e=6×10^{24} \ kg, \ R_e=6400 \ km)$

$2.05 \ hr$ | |

$1.52 \ hr$ | |

$1.75 \ hr$ | |

$1.88 \ hr$ | |

$2.15 \ hr$ |

Question 3 Explanation:

For a satellite moving in a circular orbit,

$\dfrac{m_s v^2}{r} = \dfrac{GM_e m_s}{r^2}$

$v_s=\sqrt{\dfrac{GM_e}{r}}$

Here $r=R_e+1,000 \ km$

Using $r=7.4×10^6 m$ and $M_e=6×10^{24} \ kg$

$v_s=\sqrt{\dfrac{6.67×10^{-11}×6×10^{24}}{7.4×10^6 }}=7,348 \ m/s$

Using $ T=\dfrac{2πr}{v_s}$ gives $T=6327.6 \ s=1.75 hr $

$\dfrac{m_s v^2}{r} = \dfrac{GM_e m_s}{r^2}$

$v_s=\sqrt{\dfrac{GM_e}{r}}$

Here $r=R_e+1,000 \ km$

Using $r=7.4×10^6 m$ and $M_e=6×10^{24} \ kg$

$v_s=\sqrt{\dfrac{6.67×10^{-11}×6×10^{24}}{7.4×10^6 }}=7,348 \ m/s$

Using $ T=\dfrac{2πr}{v_s}$ gives $T=6327.6 \ s=1.75 hr $

Question 4 |

### Alex drops one object from points A and B each. He then finds the ratio of acceleration due to gravity for those objects.

### Which of the below options shows the ratio obtained by Alex?

$0.54 ∶ 1$ | |

$0.49 ∶ 1$ | |

$0.45 ∶ 1$ | |

$0.60 ∶ 1$ | |

$0.36 ∶ 1$ |

Question 4 Explanation:

If $h$= Distance from the surface of the Earth then,

$g(h)=\dfrac{GM_e}{(R_e+h)^2}$

For point $A, h=2.5R_e-R_e=1.5R_e$

$g(h)=\dfrac{GM_e}{(R_e+1.5R_e )^2} =\dfrac{GM_e}{6.25R_e^2}$

For point $B, h=1.5R_e-R_e=0.5R_e$

$g(h)=\dfrac{GM_e}{(R_e+0.5R_e )^2} =\dfrac{GM_e}{2.25R_e^2}$

Ratio of acceleration $=\dfrac{GM_e}{6.25R_e^2 }×\dfrac{2.25R_e^2}{GM_e}=0.36 ∶ 1$

$g(h)=\dfrac{GM_e}{(R_e+h)^2}$

For point $A, h=2.5R_e-R_e=1.5R_e$

$g(h)=\dfrac{GM_e}{(R_e+1.5R_e )^2} =\dfrac{GM_e}{6.25R_e^2}$

For point $B, h=1.5R_e-R_e=0.5R_e$

$g(h)=\dfrac{GM_e}{(R_e+0.5R_e )^2} =\dfrac{GM_e}{2.25R_e^2}$

Ratio of acceleration $=\dfrac{GM_e}{6.25R_e^2 }×\dfrac{2.25R_e^2}{GM_e}=0.36 ∶ 1$

Question 5 |

### A satellite is launched from the surface of a planet such that it escapes the gravitational field and travels with a speed of $10 \ km/s$ thereafter. Determine the velocity with which the satellite must be launched. (Take the mass of the planet as $3×10^{26} \ kg$ and the radius as $8×10^7 \ m$.)

$22.8 \ km/s$ | |

$24.4 \ km/s$ | |

$26.9 \ km/s$ | |

$29.5 \ km/s$ | |

$21.6 \ km/s$ |

Question 5 Explanation:

Total energy of the satellite far away from the planet $=\dfrac{1}{2} mv_{away}^2$

Total energy of the satellite on the surface of the planet $=\dfrac{1}{2} mv_{esc}^2 - \dfrac{GM_p m}{R_p}$

Equating the two equations gives,

$\dfrac{1}{2} mv_{away}^2=\dfrac{1}{2} mv_{esc}^2 - \dfrac{GM_p m}{R_p}$

$v_{away}^2=v_{esc}^2-\dfrac{2GM_p}{R_p}$

Using $v_{away}=10^4 \ m/s, M_p=3×10^{26} \ kg$ and $R_p=8×10^7 m$ gives,

$v_{esc}^2=v_{away}^2+\dfrac{2GM_p}{R_p}$

$v_{esc}^2=(10^4 )^2+\dfrac{2×6.67×10^{-11}×3×10^{26}}{8×10^7}$

$v_{esc}^2=10^8+5.00×10^8$

$v_{esc}^2=6.00×10^8$

$v_{esc}=2.44×10^4 \ m/s$

$v_{esc}=24.4 \ km/s$

Total energy of the satellite on the surface of the planet $=\dfrac{1}{2} mv_{esc}^2 - \dfrac{GM_p m}{R_p}$

Equating the two equations gives,

$\dfrac{1}{2} mv_{away}^2=\dfrac{1}{2} mv_{esc}^2 - \dfrac{GM_p m}{R_p}$

$v_{away}^2=v_{esc}^2-\dfrac{2GM_p}{R_p}$

Using $v_{away}=10^4 \ m/s, M_p=3×10^{26} \ kg$ and $R_p=8×10^7 m$ gives,

$v_{esc}^2=v_{away}^2+\dfrac{2GM_p}{R_p}$

$v_{esc}^2=(10^4 )^2+\dfrac{2×6.67×10^{-11}×3×10^{26}}{8×10^7}$

$v_{esc}^2=10^8+5.00×10^8$

$v_{esc}^2=6.00×10^8$

$v_{esc}=2.44×10^4 \ m/s$

$v_{esc}=24.4 \ km/s$

Question 6 |

### The figure below shows the motion of a planet around the Sun.

### In which position does the planet travel with the fastest speed around the Sun?

$D$ | |

$C$ | |

$B$ | |

$A$ | |

None of the above |

Question 6 Explanation:

Since, the Sun-Planet system is isolated the total mechanical energy is conserved

And total energy at any point on the trajectory is given by the expression,

$TE=\dfrac{1}{2} mv^2 - \dfrac{GMm}{r} $

For the speed (or kinetic energy) to be maximum, the term $\dfrac{GMm}{r}$ should be as large as possible (so that the difference remains a constant)

$\dfrac{GMm}{r}$ is maximum when $r$ is small

Thus, the planet has the highest speed at point $C$

And total energy at any point on the trajectory is given by the expression,

$TE=\dfrac{1}{2} mv^2 - \dfrac{GMm}{r} $

For the speed (or kinetic energy) to be maximum, the term $\dfrac{GMm}{r}$ should be as large as possible (so that the difference remains a constant)

$\dfrac{GMm}{r}$ is maximum when $r$ is small

Thus, the planet has the highest speed at point $C$

Question 7 |

### What is the total mechanical energy of a satellite that is orbiting a planet at a distance of $1.5 × 10^{10} \ m$ from the center of the planet?

### (Assume the mass of the planet to be 5×10^25 kg and the mass of the satellite to be 1.2 × 10^2 \kg)

$-16.24 × 10^6 \ J$ | |

$-10.17 × 10^6 \ J$ | |

$-11.02 × 10^6 \ J$ | |

$-15.82 × 10^6 \ J$ | |

$-13.34 × 10^6 \ J$ |

Question 7 Explanation:

For a satellite moving about a planet in a circular orbit,

$v=\sqrt{\dfrac{GM_p}{r}}→KE=\dfrac{1}{2} \dfrac{ GM_p m_s}{r} $

$PE=-\dfrac{GM_p m_s}{r}$

$TE=PE+KE$

$TE=-\dfrac{1}{2} \dfrac{GM_p m_s}{r} $

Using $M_p=5×10^{25} \ kg, r=1.5×10^{10} \ m, m_s=1.2×10^2 \ kg$

$TE=-\dfrac{1}{2} \dfrac{6.67×10^{-11}×5×10^{25}×1.2×10^2}{1.5×10^{10}}$

$TE=-13.34×10^6 \ J $

$v=\sqrt{\dfrac{GM_p}{r}}→KE=\dfrac{1}{2} \dfrac{ GM_p m_s}{r} $

$PE=-\dfrac{GM_p m_s}{r}$

$TE=PE+KE$

$TE=-\dfrac{1}{2} \dfrac{GM_p m_s}{r} $

Using $M_p=5×10^{25} \ kg, r=1.5×10^{10} \ m, m_s=1.2×10^2 \ kg$

$TE=-\dfrac{1}{2} \dfrac{6.67×10^{-11}×5×10^{25}×1.2×10^2}{1.5×10^{10}}$

$TE=-13.34×10^6 \ J $

Question 8 |

### The time period of two satellites, $LI$ and $MBII$, moving around the Earth is $24 \ hours$ and $2.5 \ hours$, respectively. The distance of the satellite $LI$ is approximately $36,000 \ km$ from the center of the Earth. What is the distance between $MBII$ satellite from the center of the Earth?

$6,324 \ km$ | |

$7,204 \ km$ | |

$8,122 \ km$ | |

$7,964 \ km$ | |

$7,684 \ km$ |

Question 8 Explanation:

According to Kepler’s 3

$T^2∝r^3$

Thus,

$\left(\dfrac{T_{MBII}}{T_{LI}} \right)^2=\left( \dfrac{r_{MBII}}{r_{LI}} \right)^3$

$\left(\dfrac{2.5}{24}\right)^2= \left(\dfrac{r_{MBII}}{36,000}\right)^3$

$\left( \dfrac{r_{MBII}}{36,000}\right)^3=\dfrac{1}{92.16}$

Solving for $r_{MBII}$ gives,

$r_{MBII}=\dfrac{36,000}{4.52}=7,964 \ km$

^{rd}law,$T^2∝r^3$

Thus,

$\left(\dfrac{T_{MBII}}{T_{LI}} \right)^2=\left( \dfrac{r_{MBII}}{r_{LI}} \right)^3$

$\left(\dfrac{2.5}{24}\right)^2= \left(\dfrac{r_{MBII}}{36,000}\right)^3$

$\left( \dfrac{r_{MBII}}{36,000}\right)^3=\dfrac{1}{92.16}$

Solving for $r_{MBII}$ gives,

$r_{MBII}=\dfrac{36,000}{4.52}=7,964 \ km$

Question 9 |

### The acceleration due to gravity varies at large distances from the surface of the Earth. If an object is dropped at a distance of $h$ from the surface of the Earth, then what is the velocity of the object $v$ as a function of its displacement $x$?

### (Hint: You might find the relation $ \dfrac{dv}{dt} = v \dfrac{dv}{dx}$ helpful)

$v(x)=2 \sqrt{ \dfrac{GM}{R_e+h-x} - \dfrac{GM}{R_e+h}}$ | |

$v(x)=2 \sqrt{ \dfrac{GM}{R_e+h+x}}$ | |

$v(x)=2 \sqrt{ \dfrac{GM}{R_e+h-x}}$ | |

$v(x)=2 \sqrt{ \dfrac{GM}{R_e+h} - \dfrac{GM}{R_e+h+x}}$ | |

$v(x)=2 \sqrt{ \dfrac{GM}{R_e+h} + \dfrac{GM}{R_e+h+x}}$ |

Question 9 Explanation:

$\dfrac{dv}{dt} = \dfrac{GM}{r^2}$

Using the differential equation from the question gives,

$v \dfrac{dv}{dx} = \dfrac{GM}{r^2}$

Note: $r=R_e+(h-x)$

Solving the differential equation gives,

$\dfrac{v^2}{2}=C + \dfrac{GM}{R_e+h-x}$

At $x=0, \ v=0$ which implies $C= - \dfrac{GM}{R_e+h}$

$v(x)=2 \sqrt{ \dfrac{GM}{R_e+h-x} - \dfrac{GM}{R_e+h}}$

Using the differential equation from the question gives,

$v \dfrac{dv}{dx} = \dfrac{GM}{r^2}$

Note: $r=R_e+(h-x)$

Solving the differential equation gives,

$\dfrac{v^2}{2}=C + \dfrac{GM}{R_e+h-x}$

At $x=0, \ v=0$ which implies $C= - \dfrac{GM}{R_e+h}$

$v(x)=2 \sqrt{ \dfrac{GM}{R_e+h-x} - \dfrac{GM}{R_e+h}}$

Question 10 |

### Two planets are aligned in such a way that they always face each other while moving in their respective orbits. A small rover is to be launched from planet $A$ to planet $B$.

### What should be the speed v of the rover with which it should be launched from planet $A$ so that it reaches planet $B$ with a speed of $u = \sqrt{ \dfrac{4Gm}{R}}$?

(Note: The values inside the circles shows the mass and the radius of the planet respectively.)

$\sqrt{\dfrac{1.66Gm}{R}}$ | |

$\sqrt{\dfrac{2.41Gm}{R}}$ | |

$\sqrt{\dfrac{3.38Gm}{R}}$ | |

$\sqrt{\dfrac{2.02Gm}{R}}$ | |

$\sqrt{\dfrac{4.56Gm}{R}}$ |

Question 10 Explanation:

Total energy of rover on planet

$A=\dfrac{1}{2} m_r v^2 - \dfrac{Gmm_r}{R} - \dfrac{2Gmm_r}{51.5R}$

$=\dfrac{1}{2} m_r v^2-1.04 \dfrac{Gmm_r}{R}$

Total energy of rover on planet

$B=\dfrac{1}{2} m_r \left(\dfrac{4Gm}{R}\right)-\dfrac{Gmm_r}{51R} - \dfrac{2Gmm_r}{1.5R}$

$=2 \dfrac{Gmm_r}{R-1.35} \dfrac{Gmm_r}{R}$

$=0.65 \dfrac{Gmm_r}{R}$

Since the total mechanical energy of the rover is a constant

$\dfrac{1}{2} m_r v^2-1.04 \dfrac{Gmm_r}{R}=0.65 \dfrac{Gmm_r}{R}$

$\dfrac{1}{2} m_r v^2=1.69 \dfrac{Gmm_r}{R}$

$v^2=\dfrac{3.38Gm}{R}$

$v=\sqrt{ \dfrac{3.38Gm}{R}} $

$A=\dfrac{1}{2} m_r v^2 - \dfrac{Gmm_r}{R} - \dfrac{2Gmm_r}{51.5R}$

$=\dfrac{1}{2} m_r v^2-1.04 \dfrac{Gmm_r}{R}$

Total energy of rover on planet

$B=\dfrac{1}{2} m_r \left(\dfrac{4Gm}{R}\right)-\dfrac{Gmm_r}{51R} - \dfrac{2Gmm_r}{1.5R}$

$=2 \dfrac{Gmm_r}{R-1.35} \dfrac{Gmm_r}{R}$

$=0.65 \dfrac{Gmm_r}{R}$

Since the total mechanical energy of the rover is a constant

$\dfrac{1}{2} m_r v^2-1.04 \dfrac{Gmm_r}{R}=0.65 \dfrac{Gmm_r}{R}$

$\dfrac{1}{2} m_r v^2=1.69 \dfrac{Gmm_r}{R}$

$v^2=\dfrac{3.38Gm}{R}$

$v=\sqrt{ \dfrac{3.38Gm}{R}} $

Question 11 |

### Which of the below statements is TRUE regarding Kepler’s 2^{nd} law?

The area swept per unit time by a planet changes at every point on the orbit | |

The time taken to complete an orbit is proportional to the $1.5th$ power of the radius of the orbit | |

Kepler’s 2 ^{nd} law is a consequence of the conservation of angular momentum in central forces | |

The angular momentum of the planet changes at every point on the orbit | |

Kepler’s 2 ^{nd} law is true only for circular orbits |

Question 11 Explanation:

Option A $→$ False, because the area swept per unit time is a constant

Option B $→$ False, because the 2

Option C $→$ True, because the 2

Option D $→$ False, because the angular momentum of the planet is a constant

Option E $→$ False, because it is true for circular and elliptical orbits

Option B $→$ False, because the 2

^{nd}law of Kepler is based on rate of change of areaOption C $→$ True, because the 2

^{nd}law of Kepler can be derived using law of conservation of angular momentumOption D $→$ False, because the angular momentum of the planet is a constant

Option E $→$ False, because it is true for circular and elliptical orbits

Question 12 |

### Look at the figure below.

### What should be the perihelion speed of planet $A$?

$1.5v_A \ \cos θ$ | |

$0.5v_A \ \sin θ$ | |

$2v_A \ \cos θ$ | |

$2v_A \ \sin θ$ | |

$3.5v_A \ \tan θ$ |

Question 12 Explanation:

At point shown,

Angular momentum

$L=m_A v_A r \ \sin (π-θ)$

$=m_A v_A r \ \sin θ$

Perihelion speed is when the planet is closest to the Sun

At perihelion,

Angular momentum

$L=m_A v_A' \ (0.5r) \ \sin 90° $

$=0.5m_A v_A' r $

Thus, $v_A'=2v_A \ \sin θ$

Angular momentum

$L=m_A v_A r \ \sin (π-θ)$

$=m_A v_A r \ \sin θ$

Perihelion speed is when the planet is closest to the Sun

At perihelion,

Angular momentum

$L=m_A v_A' \ (0.5r) \ \sin 90° $

$=0.5m_A v_A' r $

Thus, $v_A'=2v_A \ \sin θ$

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