A squash ball hits a wall with an initial speed of u and it rebounds in the opposite direction with speed v. Knowing that some energy has been lost through heat in the collision so that v < u and that the duration of the collision is t, find the magnitude of the average force exerted by the wall on the ball.
a) F = m(u+v)/t
b) f = m(u-v)/t
c) F = 0
d) F = m(2v-u)/t
Let's use second Newton's law "F=ma" ,
where F and a are average values and let's select positive direction along force;
in this coordinates velocity before collision will be negative
"a=\\frac{\\Delta v}{t}=\\frac{v-(-u)}{t}=\\frac{u+v}{t}\\\\\nF=\\frac{m(u+v)}{t}"
Answer: a)
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