Question

A 3 kg cat can jump at 3 m/s. The cat rests on an 8 kg sled(A), then jumps to another 8 kg sled(B). The cat immediately spins around and jumps back to the first sled(A). The sleds are on a frictionless surface.

a. What is the final speed of sled(A)?

b. What is the final speed of sled(B)?

c. What total impulse was imparted to sled(B)?

d. What was the overall change in momentum of sled(A)?

Answer 1

Let mc is the mass of the cat.

mA is the mass of the sled A

mB is the mass of the sled B.

Vc is the velocity of the cat.

a) Initial speed of the sled A is O .

When cat jumps on sled B, due to change in momentum , the velocity of the sled A is given by

mAUA + mcVc = mAVA   

Hence VA = ((8x0) + ( 3x3) )/ 8 = 9/8 m/s

When cat jumps from Sled A to sled B , it gives a momentum of ( 3kg x 3 m/s ) to the Sled B . i.e., 9 kg m/s momentum is given to the sled B so it stats Spin.

Let UB and VB are the initial and final momentm of Sled B before and after the cat Jump.

Using law of Conservation of momentum ,

mcVc + mBUB = (mc +mB) VB

Hence Final speed of the Sled B is VB = ( mcVc+mBUB)/(mc+mB)

Hence VB = (( 3kg x 3 m/s ) + ( 8 kg x0) )/(3 kg + 8 kg ) = 9 /11 m/s = 0.81 m/s

When cat again jump on to sled A , the relative speed of the cat is given by

Vcr = ( 3 - 9/11) = 24/11 m/s

Hence the final speed of the sled A say VAF is given by

mAVA + mcVcr   = (mA+mc) VAF

(8x9/8) +( 3x24/11) = ( 8 + 3) VAF

VAF = (15.54)/11 = 1.412 m/s

b) Let UB and VB are the initial and final momentm of Sled B before and after the cat Jump.

Using law of Conservation of momentum ,

mcVc + mBUB = (mc +mB) VB

Hence Final speed of the Sled B is VB = ( mcVc+mBUB)/(mc+mB)

Hence VB = (( 3kg x 3 m/s ) + ( 8 kg x0) )/(3 kg + 8 kg ) = 9 /11 m/s = .81 m/s

c) Impulse imparted to sled B is Final momentum - initial momentum = (8x0.81) - (8x0) = 6.48 m/s

D) Overall change in momentum of A is Final momentum - initial momentum = ( 8 x1.412 m/s ) - ( 8 x 0 ) = 11.296 m/s