# Question Video: Studying the Collision of a Moving Body with a Resting Body on the Same Line Mathematics

A car 𝐴 of mass 2.5 metric tons was moving at 24 m/s in a straight line on a smooth horizontal plane. It collided with another car 𝐵, of mass 1.5 metric tons which was at rest. Directly after the impact, the velocity of car 𝐵 relative to car 𝐴 was 6 m/s. Find the actual speeds of both cars 𝑣_(𝐴) and 𝑣_(𝐵).

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### Video Transcript

A car 𝐴 of mass 2.5 metric tons was moving at 24 meters per second in a straight line on a smooth horizontal plane. It collided with another car 𝐵 of mass 1.5 metric tons which was at rest. Directly after the impact, the velocity of car 𝐵 relative to car 𝐴 was six meters per second. Find the actual speeds of both cars 𝑣 sub 𝐴 and 𝑣 sub 𝐵.

In order to answer this question, we will firstly use the conservation of momentum, which states that the momentum before is equal to the momentum after. We will use the formula 𝑚 one 𝑢 one plus 𝑚 two 𝑢 two is equal to 𝑚 one 𝑣 one plus 𝑚 two 𝑣 two, where 𝑚 one and 𝑚 two are the masses of the objects colliding. 𝑢 one and 𝑢 two are the velocities before the impact. And 𝑣 one and 𝑣 two are the velocities after the impact.

The question also includes relative velocity. And we know that the velocity of object 𝐵 relative to object 𝐴, written 𝑣 sub 𝐵𝐴, is equal to 𝑣 sub 𝐵 minus 𝑣 sub 𝐴. Let’s begin by sketching a diagram of the situation in this question.

We are told in the question that we have two cars 𝐴 and 𝐵 of masses 2.5 and 1.5 metric tons, respectively. Car 𝐴 was initially moving at 24 meters per second. Car 𝐵 was at rest, which means it has a velocity of zero meters per second. We are not told the velocities of the cars after the impact. These are 𝑣 sub 𝐴 and 𝑣 sub 𝐵 meters per second that we are trying to calculate. We are told, however, that the velocity of car 𝐵 relative to car 𝐴 is six meters per second. This means that 𝑣 sub 𝐵 minus 𝑣 sub 𝐴 is equal to six.

Adding 𝑣 sub 𝐴 to both sides of this equation, we have 𝑣 sub 𝐵 is equal to six plus 𝑣 sub 𝐴. We will call this equation one and now move on to the conservation of momentum formula.

The momentum prior to the collision is equal to 2.5 multiplied by 24 plus 1.5 multiplied by zero. And the momentum after is equal to 2.5 multiplied by 𝑣 sub 𝐴 plus 1.5 multiplied by 𝑣 sub 𝐵. The left-hand side simplifies to give us 60. And we now have a second equation involving 𝑣 sub 𝐴 and 𝑣 sub 𝐵. Clearing some space, we now have a pair of simultaneous equations that we can solve to work out 𝑣 sub 𝐴 and 𝑣 sub 𝐵.

We will begin by substituting our expression for 𝑣 sub 𝐵 in equation one into equation two. This gives us the equation 60 is equal to 2.5𝑣 sub 𝐴 plus 1.5 multiplied by six plus 𝑣 sub 𝐴. Distributing our parentheses gives us nine plus 1.5𝑣 sub 𝐴. We can then group or collect like terms on the right-hand side. This gives us 60 is equal to four 𝑣 sub 𝐴 plus nine. Subtracting nine from both sides, the left-hand side becomes 51. We can then divide through by four such that 𝑣 sub 𝐴 is equal to 12.75.

Substituting this back into equation one, we have 𝑣 sub 𝐵 is equal to six plus 12.75, which is equal to 18.75. After the collision, the speed of car 𝐴 is 12.75 meters per second and the speed of car 𝐵 is 18.75 meters per second.

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