Question

A proton (mass 1.67 × 10^-27 kg) is accelerated along a straight line by 3.6 × 10¹⁵ ms^-2. If the length covered is 3.5 cm and initial speed of proton was 2.4 × 10⁷ms^-1, the gain in kinetic energy is

• A. 1.32MeV
• B. 1.23MeV
• C. 2.13MeV
• D. 3.12MeV

Answer 1

Answer: (A)

1.32MeV

Answer 2

Using v² – u² = 2as, we get

v = (u² + 2aS)^1/2

= (2.4 × 10⁷) + 2 × 3.6 × 10¹⁵ × 0.035

= 2.88 × 10⁷ ms^-1

Chang in K.E. = $\frac { 1 } { 2 }$ m(v² – u²)

= $\frac { 1 } { 2 }$ (1.67 × 10^-27) (2.88 × 10⁷ – 2.4 × 10⁷)

= 2.11 × 10^-13 J

= 1.32 × 10⁶ eV = 1.32 MeV