A 1. 5 kg bowling pin is hit with an 8 kg bowling ball going 6. 8 m/s. The pin bounces off the ball at 3. 0 m/s. What is the speed of the bowling ball after the collision?

Answers

Answer 1

After the collision between the 1.5 kg bowling pin and the 8 kg bowling ball, the bowling ball's speed can be calculated using the law of conservation of momentum. The speed of the bowling ball after the collision is approximately 6.8 m/s.

According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, this can be represented as:

[tex]\(m_1 \cdot v_1 + m_2 \cdot v_2 = m_1 \cdot v_1' + m_2 \cdot v_2'\)[/tex]

Where:

[tex]\(m_1\)[/tex] and [tex]\(m_2\)[/tex] are the masses of the bowling pin and the bowling ball, respectively.

[tex]\(v_1\)[/tex] and [tex]\(v_2\)[/tex] are the initial velocities of the bowling pin and the bowling ball, respectively.

[tex]\(v_1'\)[/tex] and [tex]\(v_2'\)[/tex] are the final velocities of the bowling pin and the bowling ball, respectively.

Plugging in the given values, we have:

[tex]\(1.5 \, \text{kg} \cdot 6.8 \, \text{m/s} + 8 \, \text{kg} \cdot 0 \, \text{m/s} = 1.5 \, \text{kg} \cdot 3.0 \, \text{m/s} + 8 \, \text{kg} \cdot v_2'\)[/tex]

Simplifying the equation, we find:

[tex]\(10.2 \, \text{kg} \cdot \text{m/s} = 4.5 \, \text{kg} \cdot \text{m/s} + 8 \, \text{kg} \cdot v_2'\)[/tex]

Rearranging the equation to solve for [tex]\(v_2'\)[/tex], we get:

[tex]\(8 \, \text{kg} \cdot v_2' = 10.2 \, \text{kg} \cdot \text{m/s} - 4.5 \, \text{kg} \cdot \text{m/s}\) \\\(v_2' = \frac{{10.2 \, \text{kg} \cdot \text{m/s} - 4.5 \, \text{kg} \cdot \text{m/s}}}{{8 \, \text{kg}}}\)\\\(v_2' \approx 0.81 \, \text{m/s}\)[/tex]

Therefore, the speed of the bowling ball after the collision is approximately 0.81 m/s.

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Related Questions

Calculate the angular separation of two Sodium lines given as 580.0nm and 590.0 nm in first order spectrum. Take the number of ruled lines per unit length on the diffraction grating as 300 per mm?
(A) 0.0180
(B) 180
(C) 1.80
(D) 0.180

Answers

The angular separation of two Sodium lines is calculated as (C) 1.80.

The angular separation between the two Sodium lines can be calculated using the formula:

Δθ = λ/d

Where Δθ is the angular separation, λ is the wavelength difference between the two lines, and d is the distance between the adjacent ruled lines on the diffraction grating.

First, we need to convert the given wavelengths from nanometers to meters:

λ1 = 580.0 nm = 5.80 × 10⁻⁷ m
λ2 = 590.0 nm = 5.90 × 10⁻⁷ m

The wavelength difference is:

Δλ = λ₂ - λ₁ = 5.90 × 10⁻⁷ m - 5.80 × 10⁻⁷ m = 1.0 × 10⁻⁸ m

The distance between adjacent ruled lines on the diffraction grating is given as 300 lines per mm, which can be converted to lines per meter:

d = 300 lines/mm × 1 mm/1000 lines × 1 m/1000 mm = 3 × 10⁻⁴ m/line

Substituting the values into the formula, we get:

Δθ = Δλ/d = (1.0 × 10⁻⁸ m)/(3 × 10⁻⁴ m/line) = 0.033 radians

Finally, we convert the answer to degrees by multiplying by 180/π:

Δθ = 0.033 × 180/π = 1.89 degrees

Rounding off to two significant figures, the answer is:

(C) 1.80

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