Question

The net average force exerted by the surface on the disc A during impact under the assumption mg is very small compared to the impulsive force is :

• A. $\frac{m{v}_0}{\sqrt{2}}(\hat{i}+\hat{j})/△t$
• B. $-m{v}_0(\hat{i}+\hat{j})/△t$
• C. $\sqrt{2}m{v}_0(\hat{i}+\hat{j})/△t$
• D. $\sqrt{3}m{v}_0(\hat{i}+\hat{j})/△t$

Answer 1

The correct option is A $\frac{m{v}_0}{\sqrt{2}}(\hat{i}+\hat{j})/△t$

Momentum change of A $=m\frac{v_0}{\sqrt{2}}\frac{(\hat{i}+\hat{j})}{\sqrt{2}}$

$F.△T=\frac{{mv}_0}{\sqrt{2}}\frac{(\hat{i}+\hat{j})}{\sqrt{2}}F=\frac{{mv}_0}{\sqrt{2}}\frac{(\hat{i}+\hat{j})}{△T}$