Question

A 1050 kg sports car is moving westbound at 13.0 m/s on a level road when it collides with a 6320 kg truck driving east on the same road at 12.0 m/s. The two vehicles remain locked together after the collision. What is the velocity (magnitude) of the two vehicles just after the collision?

At what speed should the truck have been moving so that it and car are both stopped in the collision?
Find the change in kinetic energy of the system of two vehicles for the situations of part A.
Find the change in kinetic energy of the system of two vehicles for the situations of part C.

Answer 1

a)
`v_(3) =8.43 m/s`

b)
`v_(2)=2.15m/s`

c) ΔK=
`-28.18x10^4J`

d)ΔK=
`-10.33x10^4J`

Step-by-step explanation:

From the exercise we know that there is a collision of a sports car and a truck.

So, the sport car is going to be our object number 1 and the truck object number 2.

`m_(1)=1050kg\\v_(1)=-13m/s\\m_(2)=6320kg\\v_(2)=12m/s`

Since the two vehicles remain locked together after the collision the final mass is:

`m_(3)=7370kg`

a) To find the velocity of the two vehicles just after the collision we must use linear's momentum principle

`p_(1)=p_(2)`

`m_(1)v_(1)+ m_(2)v_(2)=m_(3)v_(3)`

`v_(3)=(m_(1)v_(1)+m_(2)v_(2))/(m_(3))=((1050kg)(-13m/s)+(6320kg)(12m/s))/(7370kg)`

`v_(3)=8.43m/s`

b) To find the speed the truck should have had so both vehicles stopped in the collision we need to use the same principle used before

`m_(1)v_(1)+ m_(2)v_(2)=0`

`v_(2)=(-m_(1)v_(1))/(m_(2) )=(-(1050kg)(-13m/s))/((6320kg))=2.15m/s`

c) To find the change in kinetic energy we need to do the following steps:

ΔK=
`k_(2)-k_(1)=(1)/(2)m_(3)v_(3)^(2)-((1)/(2)m_(1)v_(1)^(2)+(1)/(2)m_(2)v_(2)^(2) )`

ΔK=
`(1)/(2)(7370)(8.43)^(2)-((1)/(2)(1050)(-13)^(2)+(1)/(2)(6320)(12)^(2) )=-28.18x10^(4)J`

d) The change in kinetic energy where the two vehicles stopped in the collision is:

ΔK=
`k_(2)-k_(1)=0-((1)/(2)m_(1)v_(1)^(2)+(1)/(2)m_(2)v_(2)^(2) )`

ΔK=
`-((1)/(2)(1050)(-13)^(2)+(1)/(2)(6320)(2.15)^(2) )=-10.33x10^4J`