Physics, 15.11.2019 03:31 dejamjones54091

Awhite billiard ball with mass mw = 1.61 kg is moving directly to the right with a speed of v = 3.01 m/s and collides elastically with a black billiard ball with the same mass mb = 1.61 kg that is initially at rest. the two collide elastically and the white ball ends up moving at an angle above the horizontal of θw = 51° and the black ball ends up moving at an angle below the horizontal of θb = 39°.
a) what is the final speed of the white ball? (m/s)
b) what is the final speed of the black ball? (m/s)
c) what is the magnitude of the final total momentum of the system? (kg-m/s)
d) what is the final total energy of the system? (j)

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Answer from: brendacauani12345

A) 2.351 m/s

B) 1.904 m/s

C) 4.846 kgm/s

D) 7.293 J

Explanation:

According to the laws of conservation of linear momentum,

the momentum of the moving objects before impact is equal to the momentum of the objects after impact (Assuming no external forces were applied)

momentum before impact = momentum after impact

Mw* Uw + Mb * Ub = Mw * Vw + Mb * Vb

Considering the horizontal motion , →

1.61* 3.01 + 1.61 * 0 = 1.61 * (Vw)ₓ + 1.61 * (Vb)ₓ

(Vw)ₓ + (Vb)ₓ = 3.01 --------(1)

Vw*cos(51) + Vb*cos(39) = 3.01

Vw*0.63 + Vb*0.77 = 3.01 -----------------------(1)

Considering the Vertical motion , ↑

1.61* 0 + 1.61 * 0 = 1.61 * (Vw)ₙ + 1.61 * (Vb)ₙ

↑(Vw)ₓ =↓ (Vb)ₓ

↑Vw * sin(51)=↓ Vb*sin(39)

Vw= 0.81 * Vb --------------(2)

From (1) (2)

0.81 * Vb*0.63 + Vb*0.77 = 3.01

Vb = 2.351 m/s

Vw= 0.81 * 2.351

Vw= 1.904 m/s

momentum Final = momentum before

= Mw * Uw

= 1.61 *3.01

= 4.846 kgm/s

Since this is a elastic motion , no energy loss will be there

therefore , Final energy = initial energy

= 1/2 * Mw *V²w

= 1/2 * 1.61 * 3.01²

= 7.293 J

Awhite billiard ball with mass mw = 1.61 kg is moving directly to the right with a speed of v = 3.01

Answer from: Quest

the answer is b

explanation:

Answer from: Quest

5.5 n====== $\sum f = ma$ assuming that $\sum f = f_{app}$ we have m = 0.145 kg a = 38 m/s²therefore $f_{app} = (0.145\text{ kg})(38\text{ m/s}^2) = 5.51\text{ n}$ 5.51 n, but two significant figures.final answer: 5.5 n

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