Question

A 0.150-kg lump of clay is dropped from a height of 1.45 m onto the floor. It sticks to the floor and does not bounce. What is the magnitude of the impulse J imparted to the clay by the floor during the impact?

Answer 1

The magnitude of the impulse imparted to the clay by the floor during the impact is 0 Newton-seconds (N·s).

Step-by-step explanation:

The magnitude of the impulse imparted to the clay by the floor during the impact can be calculated using the principle of conservation of momentum.

Momentum is the product of mass and velocity. Since the clay sticks to the floor, its final velocity is 0 m/s. So, the initial momentum of the clay before the impact is equal to its final momentum after the impact.

Using the equation p = mv, where p is the momentum, m is the mass, and v is the velocity, we can calculate the magnitude of the impulse J imparted to the clay by the floor during the impact.

Initial momentum = Final momentum

0.150 kg * Initial velocity = 0.150 kg * 0 m/s

Initial velocity = 0 m/s

Therefore, the magnitude of the impulse J imparted to the clay by the floor during the impact is 0 Newton-seconds (N·s).

Answer 2

Approximately
`0.800\; {\rm kg\cdot m\cdot s^(-1)}`, assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

The clay in this question hits the ground right after a free fall from a given height. Given the mass of the clay and the change in the height during the free fall, the goal is to find the impulse (change in momentum) on the clay upon impact.

The impulse when the clay lands can be found in the following steps:

• Find the velocity of the clay right before landing using the SUVAT equations.
• Find the change in the velocity of the clay upon impact, and hence find the impulse on the clay.

Assuming that the air resistance on the clay is negligible. The acceleration of the clay would be constantly
`a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}` during the drop. The velocity of the clay right before landing can be found with the following equation:

`v^(2) - u^(2) = 2\, a\, x`, where:

• 
  `u` is the initial velocity,
• 
  `v` is the velocity right before landing,
• 
  `a` is the acceleration during the drop, and
• 
  `x` is the change in position during the drop.

The change in the position of the clay would be
`x = (-1.45)\; {\rm m}` (negative because the position of the clay would be below where it was dropped.) The question implies that the clay was initially stationary at the beginning of the drop, such that
`u = 0\; {\rm m\cdot s^(-1)}`. Rearrange this equation to find
`\|v\|`, the magnitude of the velocity right before landing:

`\begin{aligned}\|v\| &= \sqrt{u^(2) - 2\, a\, x} \\ &= √((0) - 2\, (-9.81)\, (-1.45))\; {\rm m\cdot s^(-1)} \\ &\approx 5.3338\; {\rm m\cdot s^(-1)}\end{aligned}`.

The value of
`v` should be negative since the clay would be travelling downward towards the ground. Therefore:

`v \approx (-5.3338)\; {\rm m\cdot s^(-1)}`.

The impulse
`J` on an object is equal to the product of mass
`m` and the change in velocity
`\Delta v`:

`J = m\, \Delta v`.

Since the clay did not bounce, the velocity of the clay would be
`0\; {\rm m\cdot s^(-1)}` after the impact. The change in the velocity of the clay would be:

`\begin{aligned}\Delta v &= (\text{final velocity}) - (\text{initial velocity}) \\ &= (0\; {\rm m\cdot s^(-1)}) - (-5.3338\; {\rm m\cdot s^(-1)}) \\ &= 5.3338\; {\rm m\cdot s^(-1)}\end{aligned}`.

Given that
`m = 0.150\; {\rm kg}`, the impulse on the clay would be:

`\begin{aligned} J &= m\, \Delta v \\ &= (0.150\; {\rm kg}) \, (5.3338\; {\rm m\cdot s^(-1)}) \\ &\approx 0.800\; {\rm kg\cdot m\cdot s^(-1)}\end{aligned}`.