A large raindrop-the type that lands with a definite splat-has a mass of 0.0014 g and hits your roof at a speed of 8.1 m/s. a. What is the magnitude of the impulse delivered to your roof? b. If the raindrop comes to rest in 0.37 ms, what is the magnitude of the force of the impact?

Answer :

Answer

given,

mass of the drop, m = 0.0014 g

speed of the drop, u = 8.1 m/s

a) Change in momentum is equal to impulse

   final velocity of the drop, v = 0 m/s

   J = m ( v - u )

   J = 0.0014 x 10⁻³ x ( 0 - 8.1 )

   J = -1.134 x 10⁻⁵ kg.m/s

impulse of the roof = - J = 1.134 x 10⁻⁵ kg.m/s

b) time, t = 0.37 m s

   impact of force = ?

  we know

     J = F x t

   1.134 x 10⁻⁵ = F x 0.37 x 10⁻³

     F =  0.031 N

the magnitude of the force of the impact is equal to F =  0.031 N

The magnitude of the impulse delivered to your roof is 1.134 x 10⁻ kg.m/s.

The magnitude of the force of the impact is 0.0307 N.

The given parameters;

  • mass of the raindrop, m = 0.0014 g = 0.0014 x 10⁻³ kg
  • speed of the rain, v = 8.1 m/s
  • time of motion, t = 0.37 ms

The impulse delivered to the roof is calculated as follows;

J = ΔP = mv

[tex]J = (0.0014\times 10^{-3})(8.1)\\\\J = 1.134 \times 10^{-5} \ kg.m/s[/tex]

The magnitude of the force of impact is calculated by applying Newton's second law of motion as follows;

F = ma

[tex]F = \frac{mv}{t} \\\\F = \frac{(0.0014\times 10^{-3})(8.1)}{0.37\times 10^{-3}} \\\\F = 0.0307 \ N[/tex]

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