# AP Physics C Mechanics: Unit 4 Practice Test — Systems of Particles & Linear Momentum

AP Physics C: Mechanics practice test 4.

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 Question 1

### Three masses, $m, 2m,$ and $3m$ are at coordinates $(a,0), (4a,0),$ and $(-a,0)$, respectively. What are the coordinates of the center of mass of the system?

 A $(a,0)$ B $(0,a)$ C $(2a,0)$ D $(3a,0)$ E $(-2a,0)$
Question 1 Explanation:
x coordinate of the center of mass

$=\dfrac{∑mx}{∑m}$

$=\dfrac{m×a+2m×4a+3m×-a}{m+2m+3m}$

$=\dfrac{6ma}{6m}$

$=a$

Since the y coordinate for each mass is $0$

The coordinates of the center of mass of the system are $(a,0)$
 Question 2

### What is the net momentum of the system of these three particles?

 A $(10.00 \skew{2.3}\hat{i} -4.00\skew{4.5}\hat{j} ) \ kgm/s$ B $(10.83 \skew{2.3}\hat{i} -6.00\skew{4.5}\hat{j} ) \ kgm/s$ C $(12.33 \skew{2.3}\hat{i} +4.29\skew{4.5}\hat{j} ) \ kgm/s$ D $(10.83 \skew{2.3}\hat{i} -3.17\skew{4.5}\hat{j} ) \ kgm/s$ E $(11.22\skew{2.3}\hat{i} -3.17\skew{4.5}\hat{j} ) \ kgm/s$
Question 2 Explanation:
Velocities of the three particles are as shown below:

$\overrightarrow{v}_1=-6\skew{5}\hat{j} \ m/s, \overrightarrow{v}_2=8\skew{2}\hat{i} \ m/s$ and

$\overrightarrow{v}_3=(4 \cos45° \skew{2}\hat{i} + 4 \sin45° \skew{5}\hat{j} ) \ m/s$

Simplifying $\overrightarrow{v}_3$ gives $(2√2 \skew{2}\hat{i} +2√2 \skew{5}\hat{j} ) \ m/s$

Momentum of each particle is equal to its velocity as $m=1 \ kg$

Net momentum $=\overrightarrow{p}_1+\overrightarrow{p}_2+\overrightarrow{p}_3$

$=(-6\skew{5}\hat{j}+8\skew{2}\hat{i}+2 \sqrt{2} \skew{2}\hat{i}+2 \sqrt{2} \skew{5}\hat{j} ) \ kgm/s$

$=(10.83\skew{2}\hat{i}-3.17\skew{5}\hat{j} ) \ kgm/s$
 Question 3

### A sphere of mass $2 \ kg$ is dropped from a height of $2 \ m$. On collision with the floor, the sphere rebounds with a speed of $3 \ m/s$. The magnitude of change in the momentum of the sphere is,

 A $9.32 \ kgm/s$ B $18.64 \ kgm/s$ C $15.24 \ kgm/s$ D $22.78 \ kgm/s$ E $20.13 \ kgm/s$
Question 3 Explanation:
Using $u=0 \ m/s$ in the equation $v^2=u^2+2gh$ gives the velocity of the sphere before collision as,

$v_b=\sqrt{2gh}$

$=\sqrt{2×10×2}$

$=6.32 m/s$ (downward)

Velocity after collision $v_a=3 \ m/s$ (upward)

Change in the velocity $=\overrightarrow{v}_a-\overrightarrow{v} b$

$=3-(-6.32)$

$=9.32 \ m/s$ (upward)

Magnitude of change in the momentum of the sphere $=2×9.32=18.64 \ kgm/s$
 Question 4

### After the collision, they combine and move with velocity $\overrightarrow{V}$. Find the magnitude of $\overrightarrow{V}$.

 A $4.34 \ m/s$ B $3.67 \ m/s$ C $11.10 \ m/s$ D $8.22 \ m/s$ E $7.55 \ m/s$
Question 4 Explanation:
Momentum of particle of mass $1.5 \ kg$ before collision

$=1.5(8 \sin ⁡60° \skew{2}\hat{i} -8 \cos ⁡60° \skew{5}\hat{j} )$

$=12(\dfrac{\sqrt{3}}{2} \skew{2}\hat{i} - \dfrac{1}{2} \skew{5}\hat{j} ) \ kgm/s$

Momentum of particle of mass $1.25 \ kg$ before collision

$=1.25(9.6 \sin60° \skew{2}\hat{i} +9.6 \cos ⁡60° \skew{5}\hat{j} )$

$=12(\dfrac{\sqrt{3}}{2} \skew{2}\hat{i} +\dfrac{1}{2} \skew{5}\hat{j} ) \ kgm/s$

Total momentum before collision
$=12\sqrt{3} \skew{2}\hat{i} \ kgm/s$

Total momentum after collision
$=2.75 \overrightarrow{V} kgm/s$

Thus, $2.75 \overrightarrow{V}=12 \sqrt{3} \skew{2}\hat{i}$

$\overrightarrow{V} =7.55\skew{2}\hat{i} \ m/s$

Speed $=V=7.55 \ m/s$
 Question 5

### The net force acting on a system of interacting particles is $0 \ N$. Which of the below statements is TRUE regarding the system?

 A The velocity of the center of mass of the system changes with time B The force between any two particles within the system is also zero C The center of mass of the system can move but its time average must be zero D The sum of all internal forces must be 0 N irrespective of the external forces E The momentum of each individual particle is conserved
Question 5 Explanation:
Option A → Velocity cannot change with time if net force is $0 \ N$

Option B → The force between any two particles may or may not be zero

Option C → The center of mass of the system can move with a fixed speed and so time average cannot be zero

Option D → The internal forces cancel in pairs and so the net internal force is $0 \ N$ and is independent of the external force(s)

Option E → The momentum of the system of particles is conserved and not for each individual particle
 Question 6

### An object moves due to force f(t) acting on it. Calculate the impulse on the object from time $t=0 \ s$ to $t=2 \ s$ if $f(t)=2e^{-t}$.

 A $2.84 \ kgm/s$ B $1.72 \ kgm/s$ C $2.46 \ kgm/s$ D $2.10 \ kgm/s$ E $1.32 \ kgm/s$
Question 6 Explanation:
$J=\int f(t) \ dt$

$=\int _0^2 2e^{-t} \ dt$

$=-2 (e^{-t}) |_0^2$

$=-2(e^{-2}-1)$

$=1.72 \ kgm/s$
 Question 7

### How far from the original line of motion does the $0.6 \ kg$ part fall?

 A $30.67 \ cm$ B $24.77 \ cm$ C $29.18 \ cm$ D $33.23 \ cm$ E $38.11 \ cm$
Question 7 Explanation:
The center of mass of the projectile falls straight down irrespective of the explosion (which is due to internal forces)

Assume the projectile is falling vertically down along the line $x=0$

$0=1×(-20)+0.6×d+2.4×16$

$0=-20+38.4+0.6d$

$d=-30.67 \ cm$

($-$ sign shows that it is on the left of the original line of motion)
 Question 8

### Which force-time graph shows the largest momentum change?

 A B C D E
Question 8 Explanation:
Area under the curve is equal to the momentum change

Option A $→$ Area $=\dfrac{1}{2}×7×6=21 \ Ns$

Option B $→$ Area $=\dfrac{1}{2}×π×3.5^2=19.24 \ Ns$

Option C $→$ Area $=3×6=18 \ Ns$

Option D $→$ Area $=\dfrac{1}{2}×3×1+1×4+\dfrac{1}{2}×1×3=7 \ Ns$

Option E $→$ Area $=\dfrac{1}{2}×(1+6)×4+1×6+\dfrac{1}{2}×6×1=23 \ Ns$

Largest momentum change is for graph E
 Question 9

### A non-uniform rod of linear density $ρ(x) = Ax^2$ (where $A$ is a positive constant) extends from $x=0$ to $x=L$. Find the center of mass of the rod.

 A $\dfrac{6}{7} \ L$ B $\dfrac{3}{5} \ L$ C $\dfrac{2}{5} \ L$ D $\dfrac{2}{3} \ L$ E $\dfrac{3}{4} \ L$
Question 9 Explanation:
Total mass $M=∫ρ(x)dx$

$=∫_0^L Ax^2 \ dx$

$=\dfrac{A}{3} (x^3) |_0^L$

$=\dfrac{A}{3} L^3$

Center of mass of the rod

$x_{cm}=\dfrac{1}{M} \int xdm$

$=\dfrac{1}{M} \int _0^L xρ(x)dx$

$=\dfrac{3}{AL^3} \int _0^L Ax^3 \ dx$

$=\dfrac{3}{AL^3} \dfrac{Ax^4}{4} |_0^L$

$=\dfrac{3}{4} \ L$
 Question 10

### Two balls of mass $m$ and $2m$, moving towards each other, suffer a head-on elastic collision. The velocities of the two balls before the collision are $-u$ and $3u$, respectively. What is the velocity of ball of mass $m$ after the collision if $2m$ continues to move in the same direction as before?

 A $\dfrac{4u}{3}$ B $\dfrac{6u}{5}$ C $\dfrac{13u}{3}$ D $-\dfrac{u}{3}$ E $\dfrac{5u}{3}$
Question 10 Explanation:
Assume ball of mass m goes to the left and $2m$ goes to the right

Total momentum before collision $=-mu+6mu=5mu$

Total kinetic energy before collision $=\dfrac{1}{2} \ mu^2+\dfrac{1}{2} \ 18mu^2=\dfrac{1}{2} \ 19mu^2$

Let the velocity of m be equal to v after the collision Speed of 2m after collision $=\dfrac{1}{2} (5u-v)$

Kinetic energy of 2m after the collision $=\dfrac{1}{8} \ 2m(5u-v)^2$

But from conservation of energy,

Kinetic energy of 2m after the collision $=\dfrac{1}{2} \ 19mu^2-\dfrac{1}{2} \ mv^2$

Solving for v gives,

$v=-u$ or $\dfrac{13u}{3}$

Rejecting $v=-u$ as it represents the original situation Thus, the velocity of m after the collision is $\dfrac{13u}{3}$
 Question 11

### If the original particle was moving to the right with a speed of $v_1$, then what should be the size of angle $y$?

 A $y = \sin^{-1} \dfrac{ \dfrac{Mv_1}{m_2 v_2} } { \left( \dfrac {m_2 v_2}{m_1 v_1} \right)^2 -1 }$ B $y = \sin^{-1} \dfrac{ \dfrac{M}{m_1} } { \left( \dfrac {m_2 v_2}{m_1 v_1} \right)^2 +1 }$ C $y = \sin^{-1} \dfrac{ \dfrac{M}{m_1} } { \left( \dfrac {m_1 v_1}{m_2 v_2} \right)^2 -1 }$ D $y = \sin^{-1} \dfrac{ \dfrac{Mv_2}{m_1 v_1} } { \left( \dfrac {m_2 v_2}{m_1 v_1} \right)^2 +1 }$ E $y = \sin^{-1} \dfrac{ \dfrac{Mv_1}{m_1 v_2} } { \left( \dfrac {m_2 v_1}{m_1 v_2} \right)^2 +1 }$
Question 11 Explanation:
From momentum conservation,

$Mv_1=m_1 v_1 \cos⁡ (90°-y) +m_2 v_2$

$\cos ⁡y→Mv_1=m_1 v_1 \sin ⁡y+m_2 v_2 \cos ⁡y … (1)$

$0=-m_1 v_1 \sin ⁡(90°-y)+m_2 v_2$

$\sin ⁡y→m_1 v_1 \cos ⁡y=m_2 v_2 \sin ⁡y … (2)$

Using $(1)$ and $(2)$,

$Mv_1=m_1 v_1 \sin ⁡y+\dfrac{(m_2 v_2 )^2}{(m_1 v_1} \sin ⁡y$

$Mv_1=\sin ⁡y \left( \dfrac{(m_2 v_2 )^2}{m_1 v_1 }+m_1 v_1 \right)$

$Mv_1=m_1 v_1 \sin ⁡y \left( \left( \dfrac{m_2 v_2}{m_1 v_1 }\right)^2+1\right)$

$\sin y = \dfrac{ \dfrac{Mv_1}{m_1 v_1} } { \left( \dfrac {m_2 v_2}{m_1 v_1} \right)^2 +1 }$

$y = \sin^{-1} \dfrac{ \dfrac{M}{m_1} } { \left( \dfrac {m_2 v_2}{m_1 v_1} \right)^2 +1 }$
 Question 12

### An isolated system of four identical interacting particles is moving steadily along the $+x$ direction. The velocity of the center of mass of the system is equal to $1 \ m/s$. If the velocity of three particles is ($2\skew{2.5}\hat{i} +2\skew{4}\hat{j}) \ m/s, (\skew{2.5}\hat{i} + 2\skew{4}\hat{j}) \ m/s,$ and $(-3\skew{2.5}\hat{i} - 5\skew{4}\hat{j}) \ m/s$, then what is the velocity of the fourth particle?

 A $(-\skew{4.5}\hat{j} ) \ m/s$ B $(\skew{2.3}\hat{i} + \skew{4.5}\hat{j} ) \ m/s$ C $(\skew{2.3}\hat{i} + 4\skew{4.5}\hat{j}) \ m/s$ D $(4\skew{2.3}\hat{i} +2 \skew{4.5}\hat{j}) \ m/s$ E $(4\skew{2.3}\hat{i} + \skew{4.5}\hat{j}) \ m/s$
Question 12 Explanation:
$M\overrightarrow{v}_{cm}=m_1 \overrightarrow{v}_1+m_2 \overrightarrow{v}_2+m_3 \overrightarrow{v}_3+m_4 \overrightarrow{v}_4$

$4m(\skew{2}\hat{i} )=m(2\skew{2}\hat{i} +2\skew{5}\hat{j} )+m(\skew{2}\hat{i} +2\skew{5}\hat{j} )$
$+m(-3\skew{2}\hat{i} -5\skew{5}\hat{j} )+m\overrightarrow{v}_4$

$4\skew{2}\hat{i} =(2\skew{2}\hat{i} +2\skew{5}\hat{j} )+(\skew{2}\hat{i} +2j ̂ )+(-3\skew{2}\hat{i} -5\skew{5}\hat{j} )+\overrightarrow{v}_4$

$\overrightarrow{v}_4=\left((-2-1+3+4) \skew{2}\hat{i} +(-2-2+5) \skew{5}\hat{j} \right)$

$\overrightarrow{v}_4=(4\skew{2}\hat{i} + \skew{5}\hat{j})$
 Question 13

### What is the value of u if the ball loses 24% of its energy due to the collision?

 A $5.92 \ m/s$ B $5.43 \ m/s$ C $4.45 \ m/s$ D $5.67 \ m/s$ E $6.11 \ m/s$
Question 13 Explanation:
Kinetic energy before collision $=\dfrac{1}{2}×0.2×6.5^2 = 4.225 \ J$

Kinetic energy after collision $=76\%$ of $4.225 J=3.211 \ J$

$u=\sqrt{2×\dfrac{KE_{after}}{m}}$

Using $KE_{after}=3.211 \ J$ and $m=0.2 \ kg$ in the above equation gives,

$u=\sqrt{\dfrac{2×3.211}{0.2}} = 5.67 \ m/s$
 Question 14

### Calculate the average force $f_{avg}$ acting on the ball during the collision with the wall if the time of contact is $0.25 \ s$.

 A $9.74 \ N$ B $9.22 \ N$ C $8.56 \ N$ D $8.21 \ N$ E $10.41 \ N$
Question 14 Explanation:
Momentum before collision $=0.2×6.50=1.30 \ kgm/s$(right)

Momentum after collision $=0.2×5.67=1.134 \ kgm/s$(left)

Change in the momentum $Δp=1.134-(-1.30)=2.434 \ kgm/s$

Average force acting on the ball $f_{avg}=\dfrac{2.434}{0.25}=9.74 \ N$
 Question 15

### If the actual variation in force is given in the graph below, then what is the value of $f_{max}$?

 A $18.92 \ N$ B $12.56 \ N$ C $9.68 \ N$ D $22.06 \ N$ E $19.47 \ N$
Question 15 Explanation:
Area under the curve = Change in the momentum of the ball

Area $= \dfrac{1}{2}×f_{max}×0.25=1/8×f_{max}$

Using $Δp=2.434 \ kgm/s$ gives,

$f_{max}=8×2.434=19.47 \ N$
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