Cart a, with a mass of 0.4 kg, travels on a horizontal air track at 6 m/s and hits cart b, which has a mass of 0.8 kg and is initially at rest. after the collision the carts stick together, and the center of mass of the two cart system has a kinetic energy of
a. 3.5 j b. 7.2 j c. 2.4 j d 1.2 je. 4.8 j

C. The two cart system has a kinetic energy of 2.4 J.

Explanation:

Hi there!

The momentum of the two cart system is conserved. That means that the momentum of the system before the collision is equal to the momentum of the system after the collision. The momentum of the system is calculated by adding the momenta of the two carts:

initial momentum of the system = final momentum of the system

pA + pB =  p (A + B)

mA · vA + mB · vB = (mA + mB) · v

Where:

pA and pB = initial momentum of carts A and B respectively.

p (A +B) = momentum of the two cart system after the collision.

mA and mB = mass of carts A and B respectively.

vA and vB = initial velocity of carts A and B.

v = velocity of the two cart system.

We have the following data:

mA = 0.4 kg

mB = 0.8 kg

vA = 6 m/s

vB = 0 m/s

Solving the equation for v:

mA · vA + mB · vB = (mA + mB) · v

0.4 kg · 6 m/s + 0.8 kg · 0 m/s = (0.4 kg + 0.8 kg) · v

2.4 kg m/s = 1.2 kg · v

v = 2.4 kg m/s / 1.2 kg

v = 2 m/s

The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where m is the mass of the object and v its speed.

Replacing with the data we have obtained:

KE = 1/2 · 1.2 kg · (2 m/s)²

KE = 2.4 J

The two cart system has a kinetic energy of 2.4 J.

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.