Lonnie pitches a baseball of mass 0.20 kg. The ball arrives at home plate with a speed of 40 m/s and is batted straight back to Lonnie with a return speed of 60 m/s. If the bat is in contact with the ball for 0.050 s what is the impulse experienced by the ball? A. 360 N.s B. 20 N.s C. 400 N.s D. 9.0 N.s

Respuesta :

Answer:

B) 20N.s is the correct answer

Explanation:

The formula for the impulse is given as:

Impulse = change in momentum

Impulse = mass × change in speed

Impulse = m × ΔV

Given:

initial speed  = 40m/s

Final speed = -60 m/s (Since the the ball will now move in the opposite direction after hitting the bat, the speed is negative)

mass = 0.20 kg

Thus, we have

Impulse = 0.20 × (40m/s - (-60)m/s)

Impulse = 0.20 × 100 = 20 kg-m/s or 20 N.s

The impulse experienced by the ball is,

[tex]\rm Impulse = 20\; Nsec[/tex]

Given :

Mass = 0.20 Kg

Initial Speed = 40 m/sec

Final Speed = -60 m/sec (minus sign shows that ball move in opposite direction)

Solution :

We know that the formula of Impulse is,

[tex]\rm Impulse = m\times \Delta V[/tex]

where, m is the mass and [tex]\rm \Delta V[/tex] is change in velocity.

[tex]\rm Impulse = 0.20\times(40-(-60))[/tex]

[tex]\rm Impulse = 20\; Nsec[/tex]

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