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

Question

A car mass of 1000 kg 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 bal of mass 40 kh horizontally at a velocity of 200 m/s which strikes the car, resulting in a perfectly inelastic collision.

A. What are the X and Y components of the cannon ball's momentum just before the collision?

B. What are the X and Y components of the car-cannon ball combined lumps momementum just after the collision?

C. What is the magnitude of the final momentum?

D. If the driver slams on the breakss when the cannon ball hits, resulting in a coefficient of kinetic friction with the road of .6, how far will the car skid before coming to a stop?

Explanation / Answer

Here ,

mass of car , m = 1000 Kg

mass of cannon ball , M = 40 Kg

A)

for the cannon ball's momentum

X component of cannon ball = 40 * 200 * cos(60)

component of cannon ball = 4000 Kg.m/s

Y -component of cannon ball = 40 * 200 * sin(60)

Y - component of cannon ball = 6928.2 Kg.m/s

B)

just after the collision , the momentum of ball and car will be added

X components of the car-cannon ball combined lump = 4000 + 1000 * 10

X components of the car-cannon ball combined lump = 14000 Kg.m/s

y - components of the car-cannon ball combined lump = 6928.2 Kg.m/s

c)

magnitude of final momentum = sqrt(14000^2 + 6928.2^2)

magnitude of final momentum = 15620.5 Kg.m/s

the magnitude of final momentum is 15620.5 Kg.m/s

D)

Here , acceleration , a = - u * g

a = - 0.6 * 9.8

a = -5.88 m/s^2

initial speed , u= 15620.5/(1000 + 40)

u = 15.02 m/s

let the distance is d

Using third equation of motion

d = u^2/(2 * g)

d = 15.02^2/(2 * 5.88)

d = 19.2 m

the car will skid 19.2 m before sstopping

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