A 1500-kg car traveling due ?-40° north of east at v2-20 m/s collides at an inte

Question

A 1500-kg car traveling due ?-40° north of east at v2-20 m/s collides at an intersection , kg minivan traveling due south at v, 20 m/s. Both vehicles become entangled and move together in one direction after the collision. The coefficient of kinetic friction between the wreck vehicles and the road is given as 0.8. Assume an isolated systemand that only frictional forces act on the wreckage after the collision. (a) Calculate the magnitude and direction of the final velocity of the tangled wreckage immediately after collision. (b) (b) Estimate the magnitude and direction of the displacement of the entangled wreckage after the collision. (c) A 5-g ball moving at 5 cm/s to the right makes an elastic collision with a 6-g ball moving to the left at 3 cm/s. Find the magnitude and direction of the final velocity of each ball after the collision. Assume an isolated system. Vlx 12x

Explanation / Answer

m1 = 1500 kg

m2 = 2500 kg

v1 = -20 j

v2 = 20*cos40i + 20*sin40 j = 15.32 i + 12.8 j

momentum before collision Pi = m1*v1 + m2*v2

momentum after collison Pf = (m1+m2)*v

from coservation of momentum

total momentum is conserved

Pf = Pi

(m1+m2)*v = m1*v1 + m2*v2

(1500+2500)*v = -(2500*20j) + (1500*(15.32i + 12.85 j) )

4000*v = 22980 i - 30725 j

v = 5.745 i - 7.68 j

magnitude = sqrt(9.575^2+0.53^2) = 9.59 m/s <<<<------------ANSWER

direction = tan^-1(-7.68/5.745) = 53.2o south of east <<<<------------ANSWER

==================================

(b)

force acting fk = uk*(m1+m2)*g

the entangled wreckage comes to rest after travelling a distance d

work done by friction Wf = -fk*d

from work energy relation = work = cahnge in KE = Kf - Ki = (1/20*(m1+m2)*v^2

-fk*d = Kf - Ki

-uk*(m1+m2)*g*d = (1/2)*(m1+m2)*(0-v^2)

0.8*9.8*d = (1/2)*9.575^2

distance d = 5.85 m <<<<------------ANSWER

direction = 53.2o south of east <<<<------------ANSWER

==========================================

(c)

ELASTIC COLLISION

m1 = 5 g                           m2 = 6 g

speeds before collision

v1i = 5 cm/s                      v2i = -3 m/s

speeds after collision

v1f = ?                         v2f = ?

initial momentum before collision

Pi = m1*v1i + m2*v2i

after collision final momentum

Pf = m1*v1f + m2*v2f

from momentum conservation

total momentum is conserved

Pf = Pi

m1*v1i + m2*v2i = m1*v1f + m2*v2f .....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*v1i^2 + 0.5*m2*v2i^2

KEf =   0.5*m1*v1f^2 + 0.5*m2*v2f^2

KEi = KEf

0.5*m1*v1i^2 + 0.5*m2*v2i^2 = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 .....(2)

solving 1&2

we get

v1f = ((m1-m2)*v1i + (2*m2*v2i))/(m1+m2)

v1f = ((5 - 6)*5 - (2*6*3))/(5+6)

v1f = -3.73 cm/s

speed of 5 g ball = 3.73 cm/s to the right

v2f = ((m2-m1)*v2i + (2*m1*v1i))/(m1+m2)

v2f = (-(6 - 5)*3 + (2*5*5))/(5+6)

v2f = 4.27 cm/s

speed of 6g ball is 4.27 cm/s to the left

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