1) Two cars collide at an intersection. Car A , with a mass of 1800 kg , is goin

Question

1) Two cars collide at an intersection. Car A, with a mass of 1800 kg , is going from west to east, while car B , of mass 1500 kg , is going from north to south at 15 m/s . As a result of this collision, the two cars become enmeshed and move as one afterwards. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 60 south of east from the point of impact.

Part A

How fast were the enmeshed cars moving just after the collision?

Part B

How fast was car A going just before the collision?

Question 2) Blocks A (mass 5.00 kg ) and B (mass 11.00 kg ) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 4.00 m/s . The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let +x be the direction of the initial motion of A.

Part A

Find the maximum energy stored in the spring bumpers.

Part B

Find the velocity of A when the maximum energy is stored in the spring bumpers.

Part C

Find the velocity of B when the maximum energy is stored in the spring bumpers.

Part D

Find the velocity of A after the blocks have moved apart.

Part E

Find the velocity of B after the blocks have moved apart.

Explanation / Answer

Q 1 )

given

the mass of the car A is mA = 1800 kg ,

mass of the car is B is mB = 1500 kg

speed of the car is B uB = 14 m/sec

and the angle is 60o  

using the conservation of momentum in y direction is

V = mB uB / ( mA + mB ) sin60

V = 1500 X 14 / ( 1500 + 1800 ) X sin60

V = 7.348 m/sec

the momemtum is in x direction

mA uA = ( mA + mB ) X V X cos60

uA = ( 1800 + 1500 ) X 7.348 X cos60 / 1800

uA = 6.7356 m/sec

Q 2 )

Part A )

momentum is of block is A

P1 = m1 v1

and momentum is of block is B

P2 = m2 u2  

Pi = P1 + P2

and Pf = ( m1 + m2 ) X Vcm  

Vcm = m1 v1 / ( m1 + m2 )

= 5 X 4 / ( 5 + 11 )

Vcm = 1.25 m/sec

the energy is

U = 1/2 m1 v12 - 1/2 ( m1 + m2 ) Vcm2

= 0.5 ( 5 X 42 - ( 5 + 11 ) 1.252 )

U = 27.5 J

Part B

the velocity of A

Vcm = 1.25 m/sec

Part C

the velocity of B

Vcm = 1.25 m/sec

Part D )

VA = ( m1 - m2 ) v1 / m1 + m2

VA = ( 5 - 11 ) X 4 / 5 + 11

VA = - 2.75 m/sec

Part E )

V2 = 2 m1 v1 / ( m1 + m2 )

V2 = 2 X 5 X 4 / ( 5 + 11 )

V2 = 2.5 m/sec

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