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Yes, you have to consider what the frame of reference is for the colliding objects.

The relative velocity between the objects is given by v = v1 - v2 where v2 is the velocity of still object and v1 is the velocity of the car.

In the first situation the relative velocity is is (50 - 0) mph. Thus the first object will hit the standing object at 50mph.

In the second situation, the relative velocity is (70 - 20) mph so again the car will hit the moving object at 50mph.

In both cases, the impact happens at the same velocity even though one is moving and one is going faster. The acceleration of the objects is the same in both situations since they both change by 50mph during the impact time with everything else kept equal.

Edit: In the elastic collision equations, you must first set the velocities relative to each other. By setting the relative velocity to 50mph in both equations, you will find that the collisions are the same.

Note:

Kinetic energy depends on the frame of reference.

If the ground is the frame of reference, then the total energy in the first situation is 1/2*50^2 = 1250 units

In the second frame of reference the total energy is 1/270^2 + 1/220^2 = 2650 units.

Assuming equal masses and an elastic collision, in the first situation the object that gets hit will move at 50mph relative to the ground. The amount of kinetic energy the object gained was 1250 units relative to the ground.

In the second situation, the impacted object now moves at 70mph, and gained 1/270^2 - 1/220^2 = 2250 units of kinetic energy relative to the ground.

From the perspective of the ground, the energy transferred to the object in the second situation is greater, but has no effect on the actual impact of the objects. The impact of the collision will still happen at the same relative velocity, so the objects in the collisions still experience the same change in momentum.

The only difference is that in order to come to a stop after the collision, it will take more energy in the second situation than the first so this may impact what forces the objects experience as they come to stop on the ground (consider frictional forces on asphalt).

Yes, you have to consider what the frame of reference is for the colliding objects.

The relative velocity between the objects is given by $v = v_1 - v_2$ where $v_2$ is the velocity of the stationary object and $v_1$ is the velocity of the car.

In the first situation the relative velocity is is $(50 - 0)\ \rm{mph}$. Thus the first object will hit the standing object at $50\ \rm{mph}$.

In the second situation, the relative velocity is $(70 - 20)\ \rm{mph}$ so again the car will hit the moving object at $50\ \rm{mph}$.

In both cases, the impact happens at the same velocity even though one is moving and one is going faster. The acceleration of the objects is the same in both situations since they both change by $50\ \rm{mph}$ during the impact time with everything else kept equal.

Edit: In the elastic collision equations, you must first set the velocities relative to each other. By setting the relative velocity to $50\ \rm{mph}$ in both equations, you will find that the collisions are the same.

Note:

Kinetic energy depends on the frame of reference.

If the ground is the frame of reference, then the total energy in the first situation is $\frac 12\cdot 50^2 = 1250\ \rm{units}$

In the second frame of reference the total energy is $\frac 12\cdot70^2 + \frac 12\cdot 20^2 = 2650\ \rm{units}$.

Assuming equal masses and an elastic collision, in the first situation the object that gets hit will move at $50\ \rm{mph}$ relative to the ground. The amount of kinetic energy the object gained was $1250\ \rm{units}$ relative to the ground.

In the second situation, the impacted object now moves at $70\ \rm{mph}$, and gained $\frac 12\cdot70^2 - \frac 12\cdot20^2 = 2250\ \rm{units}$ of kinetic energy relative to the ground.

From the perspective of the ground, the energy transferred to the object in the second situation is greater, but has no effect on the actual impact of the objects. The impact of the collision will still happen at the same relative velocity, so the objects in the collisions still experience the same change in momentum.

The only difference is that in order to come to a stop after the collision, it will take more energy in the second situation than the first so this may impact what forces the objects experience as they come to stop on the ground (consider frictional forces on asphalt).

Yes, you have to consider what the frame of reference is for the colliding objects.

The relative velocity between the objects is given by v = v1 - v2 where v2 is the velocity of still object and v1 is the velocity of the car.

In the first situation the relative velocity is is (50 - 0) mph. Thus the first object will hit the standing object at 50mph.

In the second situation, the relative velocity is (70 - 20) mph so again the car will hit the moving object at 50mph.

In both cases, the impact happens at the same velocity even though one is moving and one is going faster. The acceleration of the objects is the same in both situations since they both change by 50mph during the impact time with everything else kept equal.

Edit: In the elastic collision equations, you must first set the velocities relative to each other. By setting the relative velocity to 50mph in both equations, you will find that the collisions are the same.

Note:

Kinetic energy depends on the frame of reference.

If the ground is the frame of reference, then the total energy in the first situation is 1/2*50^2 = 1250 units

In the second frame of reference the total energy is 1/270^2 + 1/220^2 = 2650 units.

Assuming equal masses and an elastic collision, in the first situation the object that gets hit will move at 50mph relative to the ground. The amount of kinetic energy the object gained was 1250 units relative to the ground.

In the second situation, the impacted object now moves at 70mph, and gained 1/270^2 - 1/220^2 = 2250 units of kinetic energy relative to the ground.

From the perspective of the ground, the energy transferred to the object in the second situation is greater, but has no effect on the actual impact of the objects. The impact of the collision will still happen at the same relative velocity, so the objects in the collisions will feel the same effects in both situations.

The only difference is that in order to come to a stop after the collision, it will take more energy in the second situation than the first so this may impact what forces the objects experience as they come to stop on the ground (consider frictional forces on asphalt).

Yes, you have to consider what the frame of reference is for the colliding objects.

The relative velocity between the objects is given by v = v1 - v2 where v2 is the velocity of still object and v1 is the velocity of the car.

In the first situation the relative velocity is is (50 - 0) mph. Thus the first object will hit the standing object at 50mph.

In the second situation, the relative velocity is (70 - 20) mph so again the car will hit the moving object at 50mph.

In both cases, the impact happens at the same velocity even though one is moving and one is going faster. The acceleration of the objects is the same in both situations since they both change by 50mph during the impact time with everything else kept equal.

Edit: In the elastic collision equations, you must first set the velocities relative to each other. By setting the relative velocity to 50mph in both equations, you will find that the collisions are the same.

Note:

Kinetic energy depends on the frame of reference.

If the ground is the frame of reference, then the total energy in the first situation is 1/2*50^2 = 1250 units

In the second frame of reference the total energy is 1/270^2 + 1/220^2 = 2650 units.

Assuming equal masses and an elastic collision, in the first situation the object that gets hit will move at 50mph relative to the ground. The amount of kinetic energy the object gained was 1250 units relative to the ground.

In the second situation, the impacted object now moves at 70mph, and gained 1/270^2 - 1/220^2 = 2250 units of kinetic energy relative to the ground.

From the perspective of the ground, the energy transferred to the object in the second situation is greater, but has no effect on the actual impact of the objects. The impact of the collision will still happen at the same relative velocity, so the objects in the collisions still experience the same change in momentum.

The only difference is that in order to come to a stop after the collision, it will take more energy in the second situation than the first so this may impact what forces the objects experience as they come to stop on the ground (consider frictional forces on asphalt).

Yes, you have to consider what the frame of reference is for the colliding objects.

The relative velocity between the objects is given by v = v1 - v2 where v2 is the velocity of still object and v1 is the velocity of the car.

Tn the first situation the relative velocity is is (50 - 0) mph. Thus the first object will hit the standing object at 50mph.

In the second situation, the relative velocity is (70 - 20) mph so again the car will hit the moving object at 50mph.

In both cases, the impact happens at the same velocity even though one is moving and one is going faster.

Edit: In the elastic collision equations, you must first set the velocities relative to each other. By setting the relative velocity to 50mph in both equations, you will find that the collisions are the same.

Note:

Kinetic energy depends on the frame of reference.

If the ground is the frame of reference, then the total energy in the first situation is 1/2*50^2 = 1250 units

In the second frame of reference the total energy is 1/270^2 + 1/220^2 = 2650 units.

Assuming equal masses and an elastic collision, in the first situation the object that gets hit will move at 50mph relative to the ground. The amount of kinetic energy the object gained was 1250 units relative to the ground.

In the second situation, the impacted object now moves at 70mph, and gained 1/270^2 - 1/220^2 = 2250 units of kinetic energy relative to the ground.

From the perspective of the ground, the energy transferred to the object in the second situation is greater, but has no effect on the actual impact of the objects. The impact of the collision will still happen at the same relative velocity, so the objects in the collisions will feel the same effects in both situations.

The only difference is that in order to come to a stop after the collision, it will take more energy in the second situation than the first so this may impact what forces the objects experience as they come to stop on the ground (consider frictional forces on asphalt).

Yes, you have to consider what the frame of reference is for the colliding objects.

The relative velocity between the objects is given by v = v1 - v2 where v2 is the velocity of still object and v1 is the velocity of the car.

In the first situation the relative velocity is is (50 - 0) mph. Thus the first object will hit the standing object at 50mph.

In the second situation, the relative velocity is (70 - 20) mph so again the car will hit the moving object at 50mph.

In both cases, the impact happens at the same velocity even though one is moving and one is going faster. The acceleration of the objects is the same in both situations since they both change by 50mph during the impact time with everything else kept equal.

Edit: In the elastic collision equations, you must first set the velocities relative to each other. By setting the relative velocity to 50mph in both equations, you will find that the collisions are the same.

Note:

Kinetic energy depends on the frame of reference.

If the ground is the frame of reference, then the total energy in the first situation is 1/2*50^2 = 1250 units

In the second frame of reference the total energy is 1/270^2 + 1/220^2 = 2650 units.

Assuming equal masses and an elastic collision, in the first situation the object that gets hit will move at 50mph relative to the ground. The amount of kinetic energy the object gained was 1250 units relative to the ground.

In the second situation, the impacted object now moves at 70mph, and gained 1/270^2 - 1/220^2 = 2250 units of kinetic energy relative to the ground.

From the perspective of the ground, the energy transferred to the object in the second situation is greater, but has no effect on the actual impact of the objects. The impact of the collision will still happen at the same relative velocity, so the objects in the collisions will feel the same effects in both situations.

The only difference is that in order to come to a stop after the collision, it will take more energy in the second situation than the first so this may impact what forces the objects experience as they come to stop on the ground (consider frictional forces on asphalt).