Question

a 0.24 kg blob of clay is thrown at a wall with an initial velocity of 19m/s. the clay comes to a stop against the wall in 82ms, what is the average force experienced by the clay

Answer 1

As per Newton's law we can say

$F = \frac{dP}{dt}$

here we can say

$F = \frac{m(v_f - v_i)}{\delta t}$

now we know that

$v_f = 0$

$v_i = 19 m/s$

$\delta t = 82 ms$

mass = 0.24 kg

now by above formula we will have

[tex]F = \frac{0.24(0 - 19)}{82*10^{-3}}[\tex]

so force on the wall is given by

F = 55.6 N