Question

this is a 3 part question20) A 9.50-g bullet has a speed of 1.30 km>s. (a) What is its kinetic energy in joules? (b) What is the bullet’s kinetic energy if its speed is halved? (c) If its speed is doubled?

Answer 1

Given that mass of bullet, m = 9.50 g = 0.0095 kg

speed of bullet, v is 1.30 km/s

(a) Kinetic energy is given by the formula

`K\mathrm{}E\text{. = }(1)/(2)mv^2`

Substituting the values in the above formula, we get

`K\mathrm{}E\mathrm{}=(1)/(2)(0.0095)(1.30)^2=8.0275*10^(-3)\text{ J}`

(b) Speed of bullet is v/2

Sustituting this value in the formula of kinetic energy, we get

`\begin{gathered} K.E._1=\text{ }(1)/(2)m((v)/(2))^2 \\ =(1)/(2)*(0.0095)*((1.30)/(2))^2 \\ =2.0068*10^{-3\text{ }}J \end{gathered}`

(c) Speed of bulllet becomes 2v

Sustituting this value in the formula of kinetic energy, we get

`\begin{gathered} K.E._2=\text{ }(1)/(2)m(2v)^2 \\ =(1)/(2)*(0.0095)*(2*1.30)^2 \\ =0.03211J \end{gathered}`