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?

$(a,0)$ | |

$(0,a)$ | |

$(2a,0)$ | |

$(3a,0)$ | |

$(-2a,0)$ |

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 $(a,0)$

$=\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 |

### A system of three identical particles of mass $1 \ kg$ each are moving, as shown in the diagram below.

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

$(10.00 \skew{2.3}\hat{i} -4.00\skew{4.5}\hat{j} ) \ kgm/s$ | |

$(10.83 \skew{2.3}\hat{i} -6.00\skew{4.5}\hat{j} ) \ kgm/s$ | |

$(12.33 \skew{2.3}\hat{i} +4.29\skew{4.5}\hat{j} ) \ kgm/s$ | |

$(10.83 \skew{2.3}\hat{i} -3.17\skew{4.5}\hat{j} ) \ kgm/s$ | |

$(11.22\skew{2.3}\hat{i} -3.17\skew{4.5}\hat{j} ) \ kgm/s$ |

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 $

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

$9.32 \ kgm/s$ | |

$18.64 \ kgm/s$ | |

$15.24 \ kgm/s$ | |

$22.78 \ kgm/s$ | |

$20.13 \ kgm/s$ |

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 $=2×9.32=18.64 \ kgm/s$

$v_b=\sqrt{2gh}$

$=\sqrt{2×10×2}$

$=6.32 m/s$ (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 |

### Two particles of masses 1.5 kg and 1.25 kg are moving along different directions, as shown below.

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

$4.34 \ m/s$ | |

$3.67 \ m/s$ | |

$11.10 \ m/s$ | |

$8.22 \ m/s$ | |

$7.55 \ m/s$ |

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 $=V=7.55 \ m/s$

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

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

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?

The velocity of the center of mass of the system changes with time | |

The force between any two particles within the system is also zero | |

The center of mass of the system can move but its time average must be zero | |

The sum of all internal forces must be 0 N irrespective of the external forces | |

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

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 $f(t)=2e^{-t}$.

$2.84 \ kgm/s$ | |

$1.72 \ kgm/s$ | |

$2.46 \ kgm/s$ | |

$2.10 \ kgm/s$ | |

$1.32 \ kgm/s$ |

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 $

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

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

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

$=1.72 \ kgm/s $

Question 7 |

### A projectile of mass $4 \ kg$ falling vertically down explodes in mid-air. The projectile breaks into three parts, as shown below.

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

$30.67 \ cm$ | |

$24.77 \ cm$ | |

$29.18 \ cm$ | |

$33.23 \ cm$ | |

$38.11 \ cm$ |

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)

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?

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

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.

$\dfrac{6}{7} \ L$ | |

$\dfrac{3}{5} \ L$ | |

$\dfrac{2}{5} \ L$ | |

$\dfrac{2}{3} \ L$ | |

$\dfrac{3}{4} \ L$ |

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

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

$=\dfrac{1}{M} \int _0^L xρ(x)dx$

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

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

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

$=∫_0^L Ax^2 \ dx$

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

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

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?

$\dfrac{4u}{3}$ | |

$\dfrac{6u}{5}$ | |

$\dfrac{13u}{3}$ | |

$-\dfrac{u}{3}$ | |

$\dfrac{5u}{3}$ |

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 $\dfrac{13u}{3}$

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 |

### A radioactive particle disintegrates into two smaller particles, as shown in the diagram below.

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

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

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

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

$y = \sin^{-1} \dfrac{ \dfrac{Mv_2}{m_1 v_1} } { \left( \dfrac {m_2 v_2}{m_1 v_1} \right)^2 +1 }$ | |

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

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 $

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

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

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

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

$Mv_1=m_1 v_1 \cos (90°-y) +m_2 v_2$

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

$0=-m_1 v_1 \sin (90°-y)+m_2 v_2$

$ \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?

$(-\skew{4.5}\hat{j} ) \ m/s$ | |

$(\skew{2.3}\hat{i} + \skew{4.5}\hat{j} ) \ m/s$ | |

$(\skew{2.3}\hat{i} + 4\skew{4.5}\hat{j}) \ m/s$ | |

$(4\skew{2.3}\hat{i} +2 \skew{4.5}\hat{j}) \ m/s$ | |

$(4\skew{2.3}\hat{i} + \skew{4.5}\hat{j}) \ m/s$ |

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}) $

$4m(\skew{2}\hat{i} )=m(2\skew{2}\hat{i} +2\skew{5}\hat{j} )+m(\skew{2}\hat{i} +2\skew{5}\hat{j} )$

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

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

$\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 |

### Questions 13, 14, and 15 are based on the below information:

### A ball of mass $0.2 \ kg$ is moving towards a wall with a speed of $6.5 \ m/s$. It undergoes a partial inelastic collision and bounces back with a smaller velocity, as shown below.

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

$5.92 \ m/s$ | |

$5.43 \ m/s$ | |

$4.45 \ m/s$ | |

$5.67 \ m/s$ | |

$6.11 \ m/s$ |

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,

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

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 |

### A ball of mass $0.2 \ kg$ is moving towards a wall with a speed of $6.5 \ m/s$. It undergoes a partial inelastic collision and bounces back with a smaller velocity, as shown below.

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

$9.74 \ N$ | |

$9.22 \ N$ | |

$8.56 \ N$ | |

$8.21 \ N$ | |

$10.41 \ N$ |

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 $f_{avg}=\dfrac{2.434}{0.25}=9.74 \ N$

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 |

### A ball of mass $0.2 \ kg$ is moving towards a wall with a speed of $6.5 \ m/s$. It undergoes a partial inelastic collision and bounces back with a smaller velocity, as shown below.

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

$18.92 \ N$ | |

$12.56 \ N$ | |

$9.68 \ N$ | |

$22.06 \ N$ | |

$19.47 \ N$ |

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 $

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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