Question

Two frictionless lab carts start from rest and are pushed along a level surface by a constant force. Students measure the magnitude and duration of the force on each cart, as shown in the partially completed data tableabove, and calculate final kinetic energy and momentum. Which cart has a greater kinetic energy at the end of the push?

Answer 1

Answer:

The first cart has greater kinetic energy.

Explanation:

The mass of the second cart is 2kg, and its momentum is 10kg m/s ;therefore, it must be that

$(2kg)v = 10kg\: m/s$

which gives

$v = 5m/s$.

Then the kinetic energy of 3kg cart is

$K.E. = \dfrac{1}{2}mv^2$

$= \dfrac{1}{2}(2kg)(5m/s)^2 = 25J$

$\boxed{K.E = 25J}$

which is less than that of the 1kg cart ; therefore, the 1st cart has greater kinetic energy.

Answer 2

The first cart has the greater kinetic energy.

The given parameters:

• Momentum of first cart, P = 10 kgm/s
• Momentum of the second cart, P = 10 kgm/s
• Mass of the first cart, m = 1 kg
• Mass of the second cart, m = 2kg
• Kinetic energy of the first cart, K.E = 50 J
• Kinetic energy of the second cart, K.E = ?

The velocity of the second cart is calculated as follows;

P = mv

$v = \frac{P}{m} \\\\ v = \frac{10}{2} \\\\ v = 5 \ m/s$

The kinetic energy of the second cart is calculated as follows;

$K.E = \frac{1}{2} mv^2\\\\ K.E = \frac{1}{2} \times 2 \times (5)^2\\\\ K.E = 25 \ J$

Thus, the first cart has the greater kinetic energy.

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