contestada

A cue ball of mass ml 0.365 kg is shot at another billiard ball, with mass m2 055 kg, which is at rest. The cue ball has an initial speed of v = 8.5 m/s in the positive direction. Assume that the collision is elastic and exactly head-on. ? 25% Part (a) write an expression for the horizontal component of the billiard ball's velocity, vr after the collision, in terms of the other variables of the problem. -& 25% Part (b) what is this velocity, in meters per second? 25% Part (c) Write an expression for the horizontal component of the cue ball's velocity, vr, after the collision. 25% Part (d) what is the horizontal component of the cue ball's final velocity, in meters per second?

Respuesta :

Answer:

Part a)

Expression of speed of billiard ball is given as

[tex]v_r = \frac{2m_1 v}{m_1 + m_2}[/tex]

Part b)

Velocity of billiard ball is given as

[tex]v_r = 6.78 m/s[/tex]

Part c)

Expression for the speed of cue ball is given as

[tex]v_c = \frac{(m_1 - m_2) v}{m_1 + m_2}[/tex]

Part d)

Speed of cue ball is given as

[tex]v_c = -1.72 m/s[/tex]

Explanation:

Part a)

As we know that there is no external force on the system of two balls

so momentum is conserved here

so we have

[tex]m_1 v  + m_2 (0) = m_1 v_c + m_2 v_r[/tex]

also we know that the collision is elastic collision so we have

[tex]v_r - v_c = v - 0[/tex]

so we have

[tex]v_c = v_r - v[/tex]

so we have

[tex]m_1 v = m_1(v_r - v) + m_2 v_r[/tex]

[tex]v_r = \frac{2m_1 v}{m_1 + m_2}[/tex]

Part b)

Velocity of billiard ball is given as

[tex]v_r = \frac{2(0.365)(8.5)}{0.365 + 0.55}[/tex]

[tex]v_r = 6.78 m/s[/tex]

Part c)

Now the speed of cue ball is given as

[tex]v_c = v_r - v[/tex]

[tex]v_c = \frac{2m_1 v}{m_1 + m_2} - v[/tex]

[tex]v_c = \frac{(m_1 - m_2) v}{m_1 + m_2}[/tex]

Part d)

Now by above formula the speed of the cue ball is given as

[tex]v_c = \frac{0.365 - 0.55}{0.365 + 0.55}(8.5)[/tex]

[tex]v_c = -1.72 m/s[/tex]

Otras preguntas