A proton moving with the speed of 1.0 ｘ10⁷ m s⁻¹ passes through a 0.02 cm thick sheet of paper and emerges with speed of 2 ｘ10⁶ m s⁻¹. Assuming uniform deceleration, find retardation and time taken to pass through the paper. Ans. (2.4 x 10¹⁷m s⁻², 3.3 x 10⁻¹¹s ) 



Given:


Initial Velocity = `\vec v_i` = 1.0 ｘ10⁷ m s⁻¹

Final Velocity = `\vec v_i` = 2.0 ｘ10⁶ m s⁻¹

Distance = S = 0.02 cm = 0.0002 m =2.0 ｘ10⁻⁴ m



To Find:

Retardation (deceleration) = a = ?

Time taken to pass the paper = t = ?

Solution:

Using 3rd equation of motion

2aS = `\vec v_i^2` - `\vec v_f^2`

by putting the corresponding values

2a ｘ 2.0 ｘ10⁻⁴ m = `\(2.0 ｘ10⁶ m s⁻¹)^2` - `\ (1.0 ｘ10⁷ m s⁻¹)^2`

a ｘ 4.0 ｘ10⁻⁴ m = 4 ｘ10¹² m² s⁻² - 1.0 ｘ10¹⁴ m² s⁻²

a ｘ 4.0 ｘ10⁻⁴ m = 0.04 ｘ10¹⁴ m² s⁻² - 1.0 ｘ10¹⁴ m² s⁻²

a ｘ 4.0 ｘ10⁻⁴ m = - 0.96 ｘ10¹⁴ m² s⁻²

a = `\frac {- 0.96 ｘ10¹⁴ m² s⁻² }{4.0 ｘ10⁻⁴ m}`

a  = -0.24 ｘ10¹⁸ m s⁻²

or

a = -2.4 ｘ 10¹⁷ m s⁻²  ------------------Ans. (1)


Now using 1st equation of motion to find the time t

`\vec v_f^2` = `\vec v_i^2` + at

or

t = `\frac {vec v_f^2 - vec v_i^2}{a}`

by putting the corresponding values

t = `\frac { 2.0 ｘ10⁶ m s⁻¹ - 1.0 ｘ10⁷ m s⁻¹}{-2.4 ｘ 10¹⁷}`

t = `\frac { 0.2 ｘ10⁷ m s⁻¹ - 1.0 ｘ10⁷ m s⁻¹}{-2.4 ｘ 10¹⁷}`

t = `\frac { - 0.8 ｘ10⁷ m s⁻¹}{-2.4 ｘ 10¹⁷}`

t =  0.333 ｘ10¹⁰ s

or

t =  3.33 ｘ10⁻¹¹ s  ---------------Ans (2)


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