A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that is stationary. The carts stick together after the collision. What is the final kinetic energy of the system?

0.006075Joules

Explanation:

The final kinetic energy of the system is expressed as;

KE = 1/2(m1+m2)v²

m1 and m2 are the masses of the two bodies

v is the final velocity of the bodies after collision

get the final velocity using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

0.12(0.45) + 0/12(0) = (0.12+0.12)v

0.054 = 0.24v

v = 0.054/0.24

v = 0.225m/s

Get the final kinetic energy;

KE = 1/2(m1+m2)v

KE = 1/2(0.12+0.12)(0.225)²

KE = 1/2(0.24)(0.050625)

KE = 0.12*0.050625

KE = 0.006075Joules

Hence the final kinetic energy of the system is 0.006075Joules

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.

365.25 earth days are in the year of a light year