Question

A 12 kg aluminum toolbox is dropped from rest onto a large wooden beam. The box travel 0.2 m before contacting the beam. After impact, the box bounces 0.05 m above the beams surface. Approximately what impulse does the beam impart on the box?

A. 8.6 N.s

B. 12 N.s

C. 36 N.s

D. 42 N.s

Answer 1

36 N.s

Step-by-step explanation:

From energy conservation, potential energy of fall is equal to the kinetic energy of fall.

Mass = 12 kg

Height h1 = 0.2 m

Potential energy = mgh1

Where g is the acceleration due to gravity 9.81 m/s2

PE = 12 x 9.81 x 0.2 = 23.54 J

PE = KE = 23.54 = 0.5mv^2

23.54 = 0.5 x 12 x v^2

v^2 = 23.54/6 = 3.92

v1 = 1.98 m/s.

The ball then rises to h2 = 0.05

PE = mgh2 = 12 x 9.81 x 0.05 = 5.87 J

PE = KE = 0.5mv^2

5.87 = 0.5 x 12 x v^2

v^2 = 5.87/6 = 0.98

v2 = -0.99 (travel in opposite direction)

Impulse is the change of momentum

I = m(v1 - v2) = 12 (1.98 - (-0.99))

I = 12(1.98 + 0.99) = 35.65N.s

Approximately 36 N.s

Answer 2

`Imp = 35.652\,N\cdot s` (Option C)

Step-by-step explanation:

The initial and final speed of the toolbox is found by means of the Principle of Energy Conservation:

`(1)/(2)\cdot m \cdot v_(in)^(2) = m \cdot g \cdot h_(in)`

`v_(in) = \sqrt{2\cdot g \cdot h_(in)}`

`v_(in) = \sqrt{2\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot (0.2\,m)}`

`v_(in)\approx 1.981\,(m)/(s)`

`v_(out) = \sqrt{2\cdot g \cdot h_(out)}`

`v_(out) = \sqrt{2\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot (0.05\,m)}`

`v_(out)\approx 0.99\,(m)/(s)`

The impact is calculated by means of the Principle of Momentum Conservation and the Impulse Theorem:

`Imp = m\cdot(v_(out)-v_(in))`

`Imp = (12\,kg)\cdot \left[0.99\,(m)/(s) - (-1.981\,(m)/(s))\right]`

`Imp = 35.652\,N\cdot s`