Question

A proton is accelerated at 3•6×10^15ms^2 through a distance of 3•5cm. if the initial velocity is 2•4×10^9ms find the change in kinetic energy at the end of the distance. Taking mass of proton =1•6×10^-27kg

Answer 1

Answer:

20.16 × 10⁻¹⁴ J

Explanation:

Given :

A proton is accelerated at 3.6×10¹⁵ m/s²,

through a distance of 3.5cm.

If the initial velocity is 2.4×10⁹m/s.

Find the change in kinetic energy at the end of the distance.

(Taking mass of proton =1•6×10^-27kg)

Solution :

We know that,

Work done = Change in Kinetic energy = Force × Displacement = m a × s

⇒ (1.6 × 10⁻²⁷) × (3.6 × 10¹⁵) × (3.5 × 10⁻²)

⇒ (1.6 × 3.6 × 3.5) × (10⁻²⁷⁺¹⁵⁻²)

⇒ (5.76 × 3.5) × (10⁻¹⁴)

⇒ 20.16 × 10⁻¹⁴ J

(or)

Change in kinetic energy = $K_f - K_i$

⇒ $\frac{1}{2}mv^2-\frac{1}{2}mu^2$

⇒ $\frac{1}{2}m(v^2-u^2)$

We know that,

v² - u² = 2as,

So,

⇒ $\frac{1}{2}m(v^2-u^2)=\frac{1}{2}m(2as)$

⇒ $ma \times s$, which is same,.

⇒ (1.6 × 10⁻²⁷) × (3.6 × 10¹⁵) × (3.5 × 10⁻²)

⇒ (1.6 × 3.6 × 3.5) × (10⁻²⁷⁺¹⁵⁻²)

⇒ (5.76 × 3.5) × (10⁻¹⁴)

⇒ 20.16 × 10⁻¹⁴ J