A car of mass 1000kg is traveling in the x direction at 10 m/s. A cannon at an a

Question

A car of mass 1000kg is traveling in the x direction at 10 m/s. A cannon at an angle of 60 degrees to the x axis shoots a cannon ball of mass 40 kg horizontally at a velocity of 200 m/s which strikes the car, resulting in a perfectly inelastic collision

a.) what are the x and u components of the cannon balls momentum just before the collision

b.) what are the x and y components of the car-cannon ball combined lumps momentum just after the Collison

c.) what is the magnitude of the final momentum?

d.) If the driver slams on the brakes when the cannon ball hits resulting in the coefficient of kinetic friction with the road of 0.6m, how far will the car skid before coming to a stop?

Explanation / Answer

as there is no external force involved in the car, cannon system , total momentum of the system will remain constant

let i and j be unit vectors along +ve x and +ve y axis respectively

initial speed of the 1000 kg car=(10 m/s) i

then its initial momentum=mass*veloicty=(10000 kg.m/s) i

initial veloicty of the cannon in x direction=(-200*cos(60) m/s) i

=(-100 m/s) i

initial veloicty of cannon in y direction=(200*sin(60) m/s) j

=(173.21 m/s) j
hence total initial momentum of the cannon=mass*velocity

=-4000 i + 6928.2 j kg.m/s

then total initial momentum of the system=initial momentum of the car+initial momentum of the cannon

=10000 i -4000 i + 6928.2 j

=6000 i + 6928.2 j

a)x component of momentum=(-4000 kg.m/s)

y component of momentum=(6928.2 kg.m/s)

b)as momentum is conserved, final total momentum=initial total momentum

hence just after collision, momentum of car-cannon ball combined lump's momentum=6000 i +6928.2 j

hence x component=6000 kg.m/s

y component=6928.2 kg.m/s

c)magnitude of final momentum =sqrt(6000^2+6928.2^2)=9165.1 kg.m/s

d)

final speed =magnitude of final momentum/total mass of car-cannon system

=9165.1/(1000+40)=8.8126 m/s

acceleration due to friction=-friction coefficient*acceleration due to gravity

(-ve sign is due to the fact that friction opposes the motion)

=-0.6*9.8=-5.88 m/s^2

using the formula final speed^2-initial speed^2=2*acceleration*distance

==>0^2-8.8126^2=-2*5.88*distance

==>distance=8.8126^2/(2*5.88)=6.604 m

hence the car will skid for 6.604 m before coming to a stop

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