Suppose that the block B is moving with speed v just after the collision.
Now, assume the 2 block system is whole system. As we can see that there is not net external force acting on the system, therefore the linear momentum of the system must conserve.
From linear momentum conservation,
mAvAi + mBvBi = mAvAf + mAvAf .........(1)
where
vAi = Initial velocity of block A
vAf = Final velocity of Blcok A after collison
vBi = Initial velocity of block B
vBf = Final velocity of block B after collison
Using Equation (1)
6.9×4.8 + 13.8×3.2 = 6.9×v + 13.8×4
⇒ v = 3.2 m/s
(a) Velocity of 6.9 Kg Block will be 3.2 m/s
Change is Kinetic Energy= KEf - KEi = ΔKE ......... (2)
KE = ½mv²
ΔKE = [½mAvAf²+ ½mBvBf²] - [½mAvAi² + ½mBvBi²]
= - 4.416 J
(b) The change is Total KE is 4.416 J
PART (c)
If the block B ends up with 6.4 m/s. The using momentum conservation equation ...(1). The value of VBf will be 6.4
6.9×4.8 + 13.8×3.2 = 6.9×v + 13.8×6.4
v = - 1.6 m/s
This means that block A will move with a speed of 1.6 m/s in the backward direction.
Using equation (2)
ΔKE = [½mAvAf²+ ½mBvBf²] - [½mAvAi² + ½mBvBi²]
ΔKE = 141.312 J
(c) = The change is Kinetic Energy will be 141.312 J.
A 6.9 kg block with a speed of 4.8 m/s collides with a 13.8 kg block...
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