Question

A 6.25-kg bowling ball moving at 9.4 m/s collides with a 0.875-kg bowling pin, which is scattered at an angle of \Theta =83.5 degrees from the initial direction of the bowling ball, with a speed of 17.5 m/s.(Part A) Calculate the direction, in degrees, of the final velocity of the bowling ball. This angle should be measured in the same way that \Theta is.(Part B) Calculate the magnitude of the final velocity, in meters per second, of the bowling ball.

Answer 1

Answer:

A) $\phi = -14.93^o$

B) v = 9.47 m/s

Explanation:

Given:

mass of the bowling ball, m₁ = 6.25 kg

Initial velocity of the ball, v₁ = 9.4 m/s

mass of bowling pin, m₂ = 0.875 kg

velocity of the pin, v₂ = 17.5 m/s

Angle of scatter, Θ = 83.5°

a)The direction of the final velocity of the ball (∅)

$\tan\phi =- \frac{m_2v_2\sin\theta}{m_1v_1-(m_2v_2\cos\theta)}$

on substituting the values in the above equation, we get

$\tan\phi =- \frac{0.875\times 17.5\sin83.5^o}{6.25\times9.4-(0.875\times17.5\cos83.5^o)}$

or

$\tan\phi =- \frac{15.214}{57.02}$

or

$\phi = \tan^{-1}(-\frac{15.214}{57.02})$

or

$\phi = -14.93^o$

B) From the concept of conservation of momentum

we have

$m_1v\sin\phi = m_2v_2\sin\theta$

on substituting the values

we get

$v=\frac{0.875\times17.5\sin83.5^o}{6.25\times\sin(-14.89^o)}$

hence, the magnitude of the final velocity

v = 9.47 m/s

Answer 2

A) The direction of the final velocity of the bowling ball is; ∅ = -14.94°

B) The magnitude of the final velocity of the bowling ball is; v₁ = 9.44 m/s

We are given;

mass of the bowling ball; m₁ = 6.25 kg

Initial velocity of bowling ball; v₁ = 9.4 m/s

mass of bowling pin; m₂ = 0.875 kg

velocity of the pin, v₂ = 17.5 m/s

Initial velocity of pin; u₂ = 0 m/s

Angle at which bowling pin is scattered; θ = 83.5°

A) The collision is elastic and as such, we can use formula for conservation of linear momentum to get;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

The components of the pin velocity is;

v_x = 17.5 cos 83.5 = 1.981 i^ m/s

v_y = 17.5 sin 83.5 = 17.388 j^ m/s

Plugging in the relevant values into the equation for conservation of momentum to get;

(6.25 × 9.4)i^ + (0.875 × 0) = 6.25v₁ + 0.875( 1.981 i^ + 17.388 j^)

v₁ = [(6.25 × 9.4)i^ - 0.875( 1.981 i^ + 17.388 j^)]/6.25

v₁ = [(58.75 - 1.733375)i^ - 15.2154j^]/6.25

v₁ = 9.12266i^ - 2.434464j^

Thus, the angle of the velocity is;

∅ = tan⁻¹(-2.434464/9.12266)

∅ = -14.94°

B) The magnitude of the velocity will be the resultant of the vector velocity. Thus;

v₁ = √(9.12266² + (-2.434464)²)

v₁ = √89.149540442896

v₁ = 9.44 m/s

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