Inertia of a falling weight I use a standard device (penetrometer) that uses a 10 kg weight that is dropped a constant distance to cause a probe to penetrate soil. The penetration distance after 4 blows is recorded. If I were to use a 9 kg weight and increase the distance by 10% would the force being exerted be the same? 
 A: Well then we know $v=\sqrt{2gh}$ when you throw the weight. The impulse is $mv/\delta t = m\sqrt{2gh}/\delta t$ where $\delta t$ is time of contact regarded as same in both cases. Now in first case $10\sqrt{2gh}$ and in second $9\sqrt{2.2gh}$ [h*110/100= 1.1h] so clearly $10\sqrt2 >9\sqrt{2.2}$ so in the first case the probe will feel more fore hence more soil entering. And as well as the soil would offer a resistance of a then the penetration is $u^2/2g$ which would be more in first case rather than the second even after 4 blows. **** in case if you want to get at what height the 9 kg and the 10 kg make the same impact( or give the same force). then: $momentum_1=momentum_2${as $\delta t $ is same in both cases. $10\sqrt{2gh}= 9\sqrt{2g(h+x)}$  upon solving this we get $x=\frac{19h}{81}$ so for example if you make the 10 kg fall from 81 cm then 9kg has to fall from a height of 81+ (19/81)*81=100 cm to have same force on the probe. ******Now in case if your 9kg hammer falls 510mm then $h\times (1+19/81)= 510$ upon calculating for height of 10 kg i get 413.1 mm. So by making the 10 kg hammer  fall 413.1 mm would apply same force as you 9kg hammer galling from 510mm.
