A 1000kg car, travelling east at 30.0m/s, collides with a 3000kg truck, travelling north. After the collision, the vehicles stick together and the combined wreckage moves at 55.0◦ north of east. (a) What is the speed of the truck before the collision?

(b) What percentage of the initial kinetic energy of the system is lost during the collision?

1 Answer
Feb 27, 2016

(a) approx 14.28ms^-1 North
(b) approx 54.7 %

Explanation:

(a) For all collisions where no external forces are involved, the linear momentum

vec p = m cdot vec v

is conserved.

Let the car move along the x axis before the collision. Therefore truck moves along the y axis before the collision. Initial momentum is

vecp_(i nitial)=m_(car)cdot v_(car)+m_(truck)cdot v_(truck)
=1000cdot 30.0hatx+3000cdot v_(truck)haty

After the collision both (combined wreckage) move at 55.0^@ north of east with velocity v_"final". Equating the xand y components of momentum we obtain
From x component
1000cdot 30.0=4000cdot v_"final" cos55
v_"final"= (1000cdot 30.0)/(4000cdot0.57358)approx13.08ms^-1

From y component
3000cdot v_(truck)=4000v_"final" sin55

v_(truck)=(4000v_"final" sin55)/(3000)

=(4000cdot 13.08cdot 0.81915)/(3000)approx 14.28ms^-1 North

--.-.-.-.-.-.-..-.-.-.-.-.-.-.-.-..-.-.-.-.-.-.-.-.-.-
(b) Kinetic energy before collision

KE_"before" = KE_"car" + KE_"truck"
= 1/2 cdot m_"car" cdot v_"car"^2 + 1/2 cdot m_"truck" cdot v_"truck"^2
= 1/2 cdot 1000 cdot 30.0^2 + 1/2 cdot 3000 cdot 14.28^2
= 755 877.6 J

Kinetic energy after collision

KE_"after" = 1/2 cdot (m_"car" + m_"truck") cdot v_"final"^2

= 1/2 cdot 4000 cdot 13.08^2

= 342172.8 J

Now

KE_"lost"=K_"before"-KE_"after" = 755877.6 -342172.8=413704.8J
% KE_"lost" =413704.8/ 755877.6xx100approx 54.7