Question

The bird perched on the swing has a mass of 52.0 g, and the base of the swing has a mass of 153 g. Assume that the swing and bird are originally at rest and that the bird then takes off horizontally at 2.00 m/s. If the base can swing freely (i.e., without friction) around the pivot, how high will the base of the swing rise above its original level?

Answer 1

Answer: 0.024m

Explanation:

The first thing you must do is to find the speed of the swing when the bird leaves. Use conservation of momentum (p) to do this. p(i) = p(f)

P(i) = 0 (no motion) = p(f) = m*v + M*V where m&v are the bird and M&V are the wing

So V = -m*v/M = -0.052 x 2/0.153= -0.68m/s

Now apply conservation of energy. SO K bottom = 1/2*M*V^2 = U (potential) = M*g*h

h= V^2/(2*g) = 0..68^2/(2*9.8) = 0.024m

Answer 2

Answer:

h = 0.024m

Explanation:

Let's take into consideration the objects involved in the questions

Take:

Bird as Object A

Swing as Object B

Consider:

Momentum Before the bird takes off = Final Momentum after the bird takes off

MtVt = MaVa + MbVb

Where

Ma = 52g = 0.052kg

Mb = 153g = 0.153kg

Mt = Ma + Mb

= (52 + 153)g

= 205g = 0.205kg

Vt (initial Velocity) = 0

Va = Velocity of bird = 2.00m/s

Vb = Velocity of swing = ?

(0.205kg)(0) = (0.052Kg)(2m/s) + (0.153kg)(Vb)

0 = 0.104kgm/s + 0.153kgVb

Vb = -0.104kgm/s

--------------------

0.153kg

Vb = -0.679m/s = Velocity of swing

Therefore:

Energy after the bird take off (K.E) = Energy before the bird take off (P.E)

1/2M(Vb)2 =Mgh

Where M = Total mass in Kg = 0.205kg

Vb = Velocity of swing = -0.679m/s

g = 9.8m/s2

Let's make h subject of the formula

h = 1/2MVb

----------------

Mg

h = (0.205Kg)(0.697m/s)^2

-------------------------------------

(2)(0.205kg)(9.8m/s)

= 0.486m^2/s^2

-----------------------

19.6m/s^2

h = 0.024m