This is a case of inelastic collision because the objects stay in contact after the impact. The total momentum in a system is always conserved, regardless of collision type. Here, the momentum before the collision is given by the sum of the individual momentum of each car, and the momentum after the collision is given by the singular momentum of the joined vehicles.

$\displaystyle m_1v_1+m_2v_2=(m_1+m_2)v_3$

We are given the mass of each car and their initial velocities, allowing us to completely solve the left side of this equation.

$\displaystyle (1000kg)(10\frac{m}{s})+(750kg)(5\frac{m}{s})=(m_1+m_2)v_3$

$\displaystyle 13750\frac{kg*m}{s}=(m_1+m_2)v_3$

We can now use the combined mass of both cars to solve for their final velocity.

$\displaystyle 13750\frac{kg*m}{s}=(1000kg+750kg)v_3$

$\displaystyle v_3=\frac{13750\frac{kg*m}{s}}{1750kg}=7.86\frac{m}{s}$

Momentum is conserved in all situations. Here, we have an elastic collision, meaning that the objects do not stick together after the impact. Momentum is simply the product of mass and velocity, and it has the units $\displaystyle \small \frac{kg*m}{s}$.

$\displaystyle p=mv$

We can solve this problem by setting the initial and final momentum of the system equal to each other. The momentum of the system will be the sum of the momentum of the parts.

$\displaystyle m_1v_{i1}+m_2v_{i2}=m_1v_{f1}+m_2v_{f2}$

The second ball contributes no momentum initially, because its initial velocity is zero. The total initial momentum comes from the initial velocity of the larger ball.

$\displaystyle (0.1kg)(2\frac{m}{s})+(0.05kg)(0\frac{m}{s})=m_1v_{f1}+m_2v_{f2}$

$\displaystyle 0.2\frac{kg*m}{s}=m_1v_{f1}+m_2v_{f2}$

We know the mass of each ball and the final velocity of the smaller ball. Using these values, we can solve for the final velocity of the larger ball.

$\displaystyle 0.2\frac{kg*m}{s}=(0.1kg)v_{f1}+(0.05kg)(1\frac{m}{s})$

$\displaystyle 0.2\frac{kg*m}{s}=(0.1kg)v_{f1}+0.05\frac{kg*m}{s}$

$\displaystyle 0.15\frac{kg*m}{s}=(0.1kg)v_{f1}$

$\displaystyle v_{f1}=\frac{0.15\frac{kg*m}{s}}{0.1kg}=1.5\frac{m}{s}$

Momentum is given by the equation $\displaystyle p=mv$. Conservation of momentum states that $\displaystyle m_1v_1=m_2v_2$.

The initial momentum is given by the sum of the initial momentums of the two objects. Because the second mass starts from rest, it becomes irrelevant in the equation.

$\displaystyle m_1v_1=(3kg)(5\frac{m}{s})+(x)(0\frac{m}{s})=15\frac{kg*m}{s}$

After a perfectly inelastic collision, the two objects will stick together and their masses will add. The final velocity will apply to the combined mass of the two objects.

$\displaystyle m_2v_2=(3kg+xkg)(3\frac{m}{s})$

Since we know that $\displaystyle m_1v_1=m_2v_2$, we can use our values from these calculations to solve for $\displaystyle x$.

$\displaystyle (3kg+xkg)(3\frac{m}{s})=15\frac{kg*m}{s}$

$\displaystyle (3kg+xkg)=5kg$

$\displaystyle x=2kg$

The mass of the second object must be 2kg.

A perfectly inelastic collision is when the two bodies stick together at the end. At the beginning the two balls are traveling separately with individual momentum values. Using the momentum equation $\displaystyle p=mv$, we can see that ball A has a momentum of (4kg)(7m/s) to the right and ball B has a momentum of (6kg)(8m/s) to the left. The final momentum would be the mass of both balls times the final velocity, (4+6)(v_f). We can solve for v_f through conservation of momentum; the sum of the initial momentum values must equal the final momentum.

$\displaystyle (4kg)(7m/s) + (6kg)(-8m/s) = (10kg)(v_{f})$

Note: ball B's velocity is negative because they are traveling in opposite directions.

$\displaystyle 28 -48=10v_f$

$\displaystyle -20=10v_f$

$\displaystyle v_f = -2m/s$

The negative sign indicates the direction in which the two balls are traveling. Since the sign is negative and we indicated that traveling to the left is negative, the two balls must be traveling 2m/s to the left after the perfectly inelastic collision.

Kinetic energy and potential energy are interconverted. While total energy is conserved, kinetic energy is allowed to increase or decrease, provided that potential energy does the opposite. For example, as the child in collision 1 is starting at the top of a ramp, she has potential energy, but no kinetic energy. At the bottom of the ramp, all the potential energy has been turned into kinetic energy.

Momentum, on the other hand, is conserved in every collision.

Momentum is always conserved in collisions, and is equal to the product of mass and velocity.

$\displaystyle mv(before) = mv(after)$

$\displaystyle 60 kg*(5 m/s) + 43 kg*(0 m/s) = (60 kg + 43 kg)(v_f)$

$\displaystyle 60 kg * (5 m/s) = 103 kg * (v_f)$ 

$\displaystyle v_f = +2.9 m/s$

This is a reverse collision, and momentum is still conserved. The original momentum is zero.

$\displaystyle 0 = m_1v_1 + m_2v_2 = 43 kg * 8 m/s + 60kg * v_2$

$\displaystyle v_2 = -5.7 m/s$

Momentum is conserved in any collision, but kinetic energy is only conserved in elastic collisions. So, an inelastic collision has conservation of momentum, but not conservation of kinetic energy.

In an inelastic collision, kinetic energy is not conserved. In an elastic collision, kinetic energy is conserved and there is no transfer of energy to the surroundings.

Momentum is conserved regardless of the type of collision. All of the given quantities are conserved during an elastic collision.

Remember that Newton’s second law is applicable regardless of whether we are talking about kinematics, torque, electric, magnetic, or gravitational force, thus, F = ma. We know that the only force acting in free-fall in an environment where we can neglect air resistance (i.e. most of the problems on the MCAT) is gravity. The acceleration due to gravity is 9.8m/s². The sign could be + or – depending on how you decided to orient.