# Mass Dropped on Scale

When a mass is dropped onto something like a bathroom scale, the reading on the scale temporarily exceeds the actual weight of the mass. How do I explain this using forces and a force body diagram? Also, let's say instead of a mass and a scale, its just a person, a ball, and a scale. The person is standing on the scale with the ball in hand and throws it up in the air. When the person catches the ball, should the scale also read a value greater than the weight of the human and the ball combined? Is the reasoning for this the same as the mass and scale example?

Edit: Could the explanation be that at the instantaneous moment when the mass comes in contact with the scale, there is an instantaneous force caused by the impulse?

• Force = mass * acceleration. The scale can de-accelerate the mass quickly – Martin Beckett Mar 7 '18 at 4:52
• Thin of the dp/dt for of force, an impulse. It adds or subtracts during dt. – anna v Mar 7 '18 at 4:55
• Have you read about variable mass systems?? – Jnan Mar 7 '18 at 6:22

Dropping the mass onto the bathroom scale: $$F = mg + ma \tag1$$ where m is the kg mass of the mass dropped on the bathroom scale, g is gravitational acceleration, and $$a = \Delta v/\Delta t \tag2$$ where $v$ and $t$ are velocity and time. The maximum $a$ determines the maximum force indicated on the scale. The heavier the ball, the harder the scale surface and the stiffer the scale's spring, the higher $F$ will be (figure below). 