A 0.80-kg ball is thrown with a velocity of 13 m/s. The ball hit a 0.12-kg bowling pin. When the ball hits the pin, the pin goes flying forward at 6 m/s. What is the velocity of the ball after it hits the pin?

Answers

Answer 1

When the ball of mass 0.80kg hits a bowling pin of 0.12 kg, the velocity of the ball after it hits the pin = 12.1 m/s

The mass of the ball, [tex]m_1=0.80 kg[/tex]

The initial speed of the ball, [tex]u_1=13 m/s[/tex]

The mass of the bowling pin, [tex]m_2=0.12kg[/tex]

The initial speed of the bowling pin, [tex]u_2=0m/s[/tex]

The final speed of the bowling pin, [tex]v_2= 6m/s[/tex]

The final speed of the ball after collision, [tex]v_1 = ?[/tex]

According to the law of momentum conservation

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\0.80(13)+0.12(0)=0.80v_1+0.12(6)\\\\10.4+0=0.8v_1+0.72\\\\0.8v_1=10.4-0.72\\\\0.8v_1=9.68\\\\v_1=\frac{9.68}{0.8} \\\\v_1=12.1m/s[/tex]

The velocity of the ball after it hits the pin = 12.1 m/s

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tha

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General Formulas and Concepts:

Forces

Newton's 2nd Law of Motion: F = ma

F is force (in N)m is mass (in kg)a is acceleration (in m/s²)

Explanation:

Step 1: Define

Identify variables

F = 755 N

m = 15 kg

Step 2: Solve for a

Substitute in variables [Newton's 2nd Law of Motion]:                                 755 N = (15 kg)aIsolate a:                                                                                                            a = 151/3 m/s²

Topic: AP Physics 1 - Algebra Based

Unit: Forces

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