Question

An object is traveling such that it has a momentum of magnitude 23.3 kg.m/s and a kinetic energy of 262 J. Determine the following. (a) speed of the object in meters per second. (b) mass of the object in kilograms.

Answer 1

Explanation:

It is given that,

Momentum of an object, p = 23.3 kg-m/s

Kinetic energy, E = 262 J

(a) Momentum is given by, p = mv

23.3 = mv...........(1)

Kinetic energy is given by, $E=\dfrac{1}{2}mv^2$

m = mass of the object

v = speed of the object

$E=\dfrac{1}{2}\times (mv)\times v$

$262=\dfrac{1}{2}\times 23.3\times v$

v = 22.48 m/s

(2) Momentum, p = mv

$m=\dfrac{p}{v}$

$m=\dfrac{23.3\ kg-m/s}{22.48\ m/s}$

m = 1.03 Kg

Hence, this is the required solution.

Answer 2

a. The speed of the object in meters per second is 4.74 m/s.

b. The mass of the object in kilograms is 4.92 kilograms.

Give the following data:

• Momentum = 23.3 kg.m/s.

• Kinetic energy = 262 Joules.

a. To determine the speed of the object in meters per second:

How to calculate kinetic energy.

Mathematically, kinetic energy is given by this formula:

$K.E = \frac{1}{2} MV^2$

Where:

• K.E is the kinetic energy.

• M is the mass.

• V is the speed.

Simplifying further, we have:

$K.E = \frac{1}{2} \times (MV) \times V^2\\\\K.E = \frac{1}{2} \times (momentum) \times V^2\\\\262 = \frac{1}{2} \times 23.3 \times V^2\\\\524 =23.3V^2\\\\V^2 = \frac{524}{23.3} \\\\V=\sqrt{22.49}$

Speed, V = 4.74 m/s.

b. To determine the mass of the object in kilograms:

$Momentum = mass \times speed\\\\23.3 = mass \times 4.74\\\\Mass = \frac{23.3}{4.74}$

Mass = 4.92 kilograms.

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