Question

From the two expressions for impulse in the lab, solve for the general form of Newton's 2^nd law. rightarrow 240: use differentials instead of delta's You drop a 1 kg mass from a height of 1m. It hits the ground and doesn't bounce. Find P_c, J, if collision time is 1 second, find F.

Answer 1

the impluse is 4.43 kg m/s

Force = 4.43 N

Step-by-step explanation:

Data are given in the question

Ground = v

a = g

s = 1

As we know that

`v^2 = a^2 + 2as\\\\ = 0 + 2(g) (1)`

Therefore v =
`√(19.6)`

Now momentum before the collision is

= mv

`= 1 (√(19.6) ) (-J)`

Now momentum after collision is 0 as there is no bouncing of the ball could be done

J = Impluse = Change in momentum

`J = \Delta P = 0 - (-√(19.6 J)) \\\\ = √(19.6J)`

= 4.43

Now

`\Delta T = 1\ second \\\\ F = (\Delta P )/(\Delta T) \\\\ = √(19.6)`

= 4.43 N

Therefore the impluse is 4.43 kg m/s

Force = 4.43 N