Question

A proton (mass m = 1.67 × 10-27 kg) is being accelerated along a straight line at 2.50 × 1012 m/s2 in a machine. If the proton has an initial speed of 1.60 × 104 m/s and travels 3.90 cm, what then is (a) its speed

(b) the increase in its kinetic energy?

Answer 1

(A) Speed will be
`44.18* 10^4m/sec`

(b) Change in kinetic energy =
`1560* 10^(-19)`

Step-by-step explanation:

We have given mass of proton
`m=1.67* 10^(-27)kg`

Acceleration of the proton
`a=2.50* 10^(12)m/sec^2`

Initial velocity u =
`1.60* 10^4` m/sec

Distance traveled by proton s = 3.90 cm = 0.039 m

(a) From third equation of motion we know that

`v^2=u^2+2as`

`v^2=(1.60* 10^4)^2+2* 2.5* 10^(12)* 0.039`

`v=44.18* 10^4m/sec`

(b) Initial kinetic energy
`KE_I=(1)/(2)mv^2=(1)/(2)* 1.67* 10^(-27)* (1.6* 10^4)^2`

Final kinetic energy
`KE_F=(1)/(2)mv^2=(1)/(2)* 1.67* 10^(-27)* (44.18* 10^4)^2`

So change in kinetic energy
`\Delta KE=KE_F-KE_I=(1)/(2)* 1.6* 10^(-27)* 10^8* (44.18^2-1.6^2)=1560* 10^(-19)J`