Key Equations
\begin{align*} \sum p_{\text{initial}} = \sum p_{\text{final}} \; \; \text{The total momentum does not change in closed systems}\end{align*}
- In an inelastic collision, momentum is conserved within the system but kinetic energy is not; energy from the collision is turned into other forms besides kinetic energy. For example, some of the kinetic energy goes into producing heat and deforming the objects.
- In a perfectly inelastic collision, the objects will stick together and combine into a single mass. This means that when you calculate the final momentum of the system, you should consider there to be only one object with a larger total mass.
- Remember: you can only use conservation of momentum to solve a problem involving an inelastic collision
Example 1
Question: Two blocks collide on a frictionless surface. Afterwards, they have a combined mass of 10kg and a speed of \begin{align*}2.5\mathrm{m/s}\end{align*}. Before the collision, block A, which has a mass of 8.0kg, was at rest. What was the mass and initial speed of block B?
Solution: To find mass of block B we have a simple subtraction problem. We know that the combined mass is 10kg and the mass of block A is 8.0kg. \begin{align*} 10\mathrm{kg}-8.0\mathrm{kg}=2.0\mathrm{kg} \end{align*}
Now that we know the mass of both blocks we can find the speed of block B. We will use conservation of momentum. This was a completely inelastic collision. We know this because the blocks stuck together after the collision. This problem is one dimensional, because all motion happens along the same line. Thus we will use the equation \begin{align*} (m_A+m_B)v_f=m_A\times v_A+m_B\times v_B \end{align*} and solve for the velocity of block B. \begin{align*} (m_A+m_B)v_f=m_A\times v_A+m_Bv_B \Rightarrow \frac{(m_A+m_B)(v_f)-(m_A)(v_A)}{m_B}=v_B \end{align*}
Watch this Explanation
Simulation
Note: move the elasticity meter to 0% for perfectly inelastic collisions.
Collision Lab (PhET Simulation)
Car Collision (CK-12 Simulation)
Explore More
- Two blocks collide on a frictionless surface, as shown above. Afterwards, they have a combined mass of \begin{align*}10 \;\mathrm{kg}\end{align*} and a speed of \begin{align*}2.5 \;\mathrm{m/s}\end{align*}. Before the collision, one of the blocks was at rest. This block had a mass of \begin{align*}8.0 \;\mathrm{kg}\end{align*}. What was the mass and initial speed of the second block?
- In the above picture, the carts are moving on a level, frictionless track. After the collision all three carts stick together. Determine the direction and speed of the combined carts after the collision.
- The train engine and its four boxcars are coasting at \begin{align*}40 \;\mathrm{m/s}\end{align*}. The engine train has mass of \begin{align*}5,500 \;\mathrm{kg}\end{align*} and the boxcars have masses, from left to right, of \begin{align*}1,000 \;\mathrm{kg}\end{align*}, \begin{align*}1,500 \;\mathrm{kg}\end{align*}, \begin{align*}2,000 \;\mathrm{kg},\end{align*} and \begin{align*}3,000 \;\mathrm{kg}\end{align*}. (For this problem, you may neglect the small external forces of friction and air resistance.)
- What happens to the speed of the train when it releases the last boxcar? (Hint: Think before you blindly calculate.)
- If the train can shoot boxcars backwards at \begin{align*}30 \;\mathrm{m/s}\end{align*}relative to the train’s speed, how many boxcars does the train need to shoot out in order to obtain a speed of \begin{align*}58.75 \;\mathrm{m/s}\end{align*}?
- In Sacramento a \begin{align*}4000 \;\mathrm{kg}\end{align*} SUV is traveling \begin{align*}30 \;\mathrm{m/s}\end{align*} south on Truxel crashes into an empty school bus, \begin{align*}7000 \;\mathrm{kg}\end{align*}traveling east on San Juan. The collision is perfectly inelastic.
- Find the velocity of the wreck just after collision
- Find the direction in which the wreck initially moves
- Manrico \begin{align*}(80.0\;\mathrm{kg})\end{align*} and Leonora \begin{align*}(60.0\;\mathrm{kg})\end{align*} are figure skaters. They are moving toward each other. Manrico’s speed is \begin{align*}2.00\;\mathrm{m/s}\end{align*}; Leonora’s speed is \begin{align*}4.00\;\mathrm{m/s}\end{align*}. When they meet, Leonora flies into Manrico’s arms.
- With what speed does the entwined couple move?
- In which direction are they moving?
- How much kinetic energy is lost in the collision?
Answers to Selected Problems
- \begin{align*}2.0 \;\mathrm{kg}, 12.5 \;\mathrm{m/s}\end{align*}
- \begin{align*}0.13 \;\mathrm{m/s}\end{align*} to the left
- a. no change b. the last two cars
- a.\begin{align*} 15 \;\mathrm{m/s}\end{align*} b. \begin{align*}49^\circ \;\mathrm{S}\end{align*} of \begin{align*}\;\mathrm{E}\end{align*}
- a. 0.57 m/s b. the direction Leonora was originally travelling c. 297.26 J