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A pitcher claims he can throw a 0.145-kg baseball with as much momentum as a 3.00-g bullet moving with a speed of 1.50 x 103 m/s. What must the baseball’s speed be if the pitcher’s claim is valid?

Respuesta :

ANSWER

3.10 x 10⁴ m/s

EXPLANATION

Given:

• The mass of the baseball. m₂ = 0.145 g

,

• The mass of the bullet, m₁ = 3.00 g

,

• The speed of the bullet, v₁ = 1.50 x 10³ m/s

Find:

• The speed of the ball, v₂, if it has the same momentum as the bullet.

The momentum of an object of mass m moving at a speed v is,

[tex]p=mv[/tex]

So, in this problem, if both objects have the same momentum,

[tex]m_1v_1=m_2v_2[/tex]

Solving for v₂,

[tex]v_2=\frac{m_1v_1}{m_2}[/tex]

Replace the known values and solve,

[tex]v_2=\frac{3.00\text{ }g\cdot1.50\times10^3\text{ }m/s}{0.145\text{ }g}\approx3.10\times10^4m/s[/tex]

Hence, the speed of the baseball must be 3.10 x 10⁴ m/s.

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