I need help with parts of this Q : \" I have done 1,2 and 3) 1- What is the velo

Question

I need help with parts of this Q : " I have done 1,2 and 3)

1- What is the velocity of the center of mass of the system?

0.868

2- What is the initial velocity of car 1 in the center-of-mass reference frame?

3.6

3- What is the final velocity of car 1 in the center-of-mass reference frame?

-3.6

4- What is the final velocity of car 1 in the ground (original) reference frame?

??

5- What is the final velocity of car 2 in the ground (original) reference frame?

??

6- In a new (inelastic) collision, the same two bumper cars with the same initial velocities now latch together as they collide.

What is the final speed of the two bumper cars after the collision?

??

7- Compare the loss in energy in the two collisions:

?KEelastic| = |?KEinelastic|

|?KEelastic| > |?KEinelastic|

|?KEelastic| < |?KEinelastic|

I need help with 4,5,6 and 7

Explanation / Answer

1. Vcm = (m1v2 + m2v2) / (m1 + m2)

Vcm = [ (106 x 4.5) + (97 x -3.1) ] / (106 + 97)

Vcm = 0.868 m/s to the right

2. V1 wrt cm = v1 - Vcm

= 4.5 - 0.868 = 3.63 m/s

3. v2 wrt cm = v3 - Vcm

= -31. - 0.868 = 3.97 m/s

4. for elastic collision,

velocity of aproach = velocity of separation

4.5 + 3.1 = v2f + v1f

v1f + v2f = 7.6 .......(i)

applying momentum conservtion,

106 x 4.5 + (97 x -3.1) = (106 x -v1f) + (97 x v2f)

- 106v2f + 97v2f = 176.3 ...........(ii)

solving ,

v1f = 2.76 m/s

v2f = 4.84 m/s

Ans . v1f = 2.76 m/s

5. v2f = 4.84 m/s

6. for Inelastic collision,

both will move with same velocity after collision,

Applying momentum conservation,

106x 4.5 - 97 x 3.1 = (107 + 97)v

v = 0.864 m/s

7. there is zero loss in elastic collision.

Kelastic < Kinelastic

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