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?

(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#