Dozens of momentum problems with descriptive answers are presented to deepen your understanding of the definition of momentum and applying its formula in various situations.

**Problem (1): What is the definition of momentum in physics? **

**Solution:** the product of a particle's mass and velocity in physics is called the momentum of a particle, p=mv. Momentum is a vector quantity like velocity, acceleration, and force. The units of momentum are kg.m/s.

**Practice Problem (2): If a truck has a mass of 1000 kilograms and is traveling with a speed of v = 50 m/s, what is its momentum?**

**Solution:** By definition, momentum is the product of mass and velocity with formula $P=m\,v$. Thus, we have \begin{align*} P&=mv\\&=1000\times 50\\&=50,000\quad {\rm kg.m/s}\end{align*}

**Problem (3): If a car has a mass of 2,000 kilograms and traveling with a velocity of v = 72 km/h, what is its momentum?**

**Solution:** momentum is mass times velocity, $p=mv$ and its units in SI is $kg.m/s$. In this problem, first convert the units of velocity to the SI units i.e. ${\rm km/h \Rightarrow m/s}$ by multiplying it in the $\frac {10}{36}$. So, the momentum of the car gets as \begin{align*} P&=mv\\&=2000\times \Big(72\times \frac{10}{36}\Big)\\&=40,000\quad {\rm kg.m/s}\end{align*}

**Problem (4): An 8-kilogram bowling ball is rolling in a straight line toward you. If its momentum is 16 kg·m/sec, how fast is it traveling?**

**Solution:** using momentum formula $p=mv$ and solving for the velocity, we have \begin{align*} v&=\frac pm\\&=\frac{16}{8}\\&=2\quad {\rm m/s} \end{align*}

**Problem (5): A toy dart gun generates a dart with a momentum of 140 kg.m/s and a velocity of 4 m/s. What is the mass of the dart in grams? **

**Solution:** by applying momentum formula $p=mv$ and solving for $m$, we get \begin{align*} m&=\frac pv\\&=\frac {140}{4}\\&=35\quad {\rm kg} \end{align*} to convert it to the grams, multiply it by 1000 so the dart's mass is $35,000\,{\rm g}$.

**Problem (6): A beach ball is rolling in a straight line toward you at a speed of 0.5 m/sec. Its momentum is 0.25 kg·m/sec. What is the mass of the beach ball?**

**Solution:** the mass of the ball is given by \begin{align*} m&=\frac pv\\&=\frac {0.25}{0.5}\\&=0.5\quad {\rm kg} \end{align*}

**Problem (7): A 5,000-kilogram truck travels in a straight line with a speed of 54 km/h. What is its momentum?**

**Solution:** the SI units of momentum is $kg.m/s$, so first convert the units of speed into the SI units by multiplying it by $\frac {10}{36}$. Therefore, the truck's speed is $v=54\times \frac {10}{36}=15\,{\rm m/s}$. Consequently, the momentum is \begin{align*} p&=mv\\&=5000\times 15\\&=75,000\quad {\rm kg.m/s}\end{align*}

**Problem (8): A 1,400-kilogram car is also traveling in a straight line. Its momentum is equal to that of the truck in the previous question. What is the velocity of the car?**

**Solution**: the car's velocity is \begin{align*} v&=\frac pm\\&=\frac{75,000}{1400}\\&=53.6\quad {\rm m/s}\end{align*}

**Problem (9): A bus traveling at speed of 50 km/h has a momentum of 180,345 kg.m/s. What is the mass of the bus?**

**Solution**: first convert the speed's units to the SI units of velocity ($\rm \frac ms$) by multiplying it by $\frac {10}{36}$. Next, using the formula of momentum $p=mv$ and solving for the mass, we get \begin{align*} m&=\frac pv\\&=\frac{180,345}{50\times \frac{10}{36}}\\&=12984.84\quad {\rm kg}\end{align*}

**Problem (10): Which would take more force to stop in 10 seconds: an 8.0-kilogram ball rolling in a straight line at a speed of 0.2 m/sec or a 4.0-kilogram ball rolling along the same path at a speed of 1.0 m/sec?**

**Solution**: Newton's second law of motion says that the change in the momentum of an object is proportional to the force. Consequently, the higher the momentum of an object, the more force is needed to stop it.

$8\,{\rm kg}$-ball has a momentum of $p=mv=(8)(0.2)=1.6\,{\rm kg.m/s}$. In contrast, momentum of $4\,{\rm kg}$-ball is $p=mv=(4)(1)=4\,{\rm kg.m/s}$. Thus, the more force is needed to stop the second ball.

**Problem (11): The momentum of a truck traveling in a straight line at 15 m/sec is 41,500 kg·m/sec. What is the mass of the truck?**

**Solution**: using the momentum formula we have \begin{align*} m&=\frac pv\\&=\frac {41,500}{15}\\&=2766.6\quad {\rm kg}\end{align*}

**Problem (12): The parking brake on a 1500 kg automobile has broken, and the car has reached a momentum of 7500 kg.m/s. What is the velocity of the vehicle?**

**Solution**: using the definition of momentum and solving for the velocity, we have \begin{align*} v&=\frac pm\\&=\frac{7500}{1500}\\&=5\quad{\rm m/s}\end{align*}

**Problem (13): A proton with mass of $1.67\times 10^{-27}\,{\rm kg}$ moving at the speed of $5\times 10^{5}\,{\rm m/s}$. What is its momentum?**

**Solution**: momentum is mass multiply by velocity so \begin{align*} p=&mv\\&=(1.67\times 10^{-27})(5\times 10^{5})\\&=8.35\times 10^{-23}\quad {\rm kg.m/s}\end{align*}

**Problem (14): A 0.14-kilogram baseball is thrown in a straight line at a velocity of 30 m/sec. What is the momentum of baseball?**

**Solution**: baseball's momentum, $p=mv$, is determined as \begin{align*} p&=mv\\&=(0.14)(30)\\&=4.2\quad {\rm kg.m/s}\end{align*}

**Problem (15): A pitcher has an ability to throw a 0.14-kg baseball with the same momentum of a 3-g bullet moving at speed of 3,000 m/s. What is baseball's speed? **

**Solution**: since the momentum of the bullet $p_2=m_2v_2$ is the same as the baseball's one $p_1=m_1v_1$, so equating those and solving for the unknown speed of the baseball, we get \begin{align*} p_1&=p_2\\m_1v_1&=m_2v_2\\\Rightarrow v_2&=\frac{m_1 v_1}{m_2}\\&=\frac{0.003\times 3000}{0.14}\\&=64.28\quad {\rm m/s}\end{align*} in above we converter the bullet's mass to the SI unit of mass $kg$ by dividing it by $1000$.

**Problem (16): Another pitcher throws the same baseball in a straight line. Its momentum is 2.1 kg·m/sec. What is the velocity of the ball?**

**Solution**: using momentum formula, $p=mv$ and solving for the velocity, we have \begin{align*} v&=\frac pm\\&=\frac {2.1}{0.14}\\&=15\quad{\rm m/s} \end{align*}

**Problem (17): A 1-kilogram turtle crawls in a straight line at a speed of 0.01 m/sec. What is the turtle’s momentum?**

**Solution**: momentum is defined as the product of mass and velocity, so \begin{align*} p&=mv\\&=(1)(0.01)\\&=0.01\quad {\rm kg.m/s} \end{align*}

**Practice Problem (18): A bicycle and its rider have a mass of 100 kg. At what speed do they have the same momentum as an 1800-kg car traveling at 2 m/s?**

**Solution**: the momenta of bicycle and car are equal so we have \begin{align*} p_c&=p_b\\m_b\,v_b &= m_c\, v_c \\(100)v_b &= (1800)(10) \\ \Rightarrow v_b &= \frac{1800\times 2}{100}\\&=36\quad {\rm m/s}\end{align*}

To solve introductory momentum problems and questions, you must first learn the definition of momentum as the product of mass and velocity and then apply it to find the unknown. All the above problems are easy and could be solved without any difficulty.

**Date Created:** 10/26/2020

**Author**: PhysicsExams

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