The change in momentum of the hammer head is -1.2 kg m/s. The impulse given to the nail is -1.2 kg m/s. The average force between the hammer and the nail is -800 N.
(a) The change in momentum of the hammer head can be calculated using the formula:
Change in Momentum = Mass x Final Velocity - Mass x Initial Velocity
Substituting the given values, the change in momentum is:
Change in Momentum = 0.15 kg x 0 m/s - 0.15 kg x 8.0 m/s = -1.2 kg m/s
(b) The impulse given to the nail can be calculated using the formula:
Impulse = Force x Time
Since the hammer head stops after hitting the nail, the impulse given to the nail would be equal to the change in momentum of the hammer head, which is -1.2 kg m/s.
(c) The average force between the hammer and the nail can be calculated using the formula:
Average Force = Impulse / Time
Substituting the given values, the average force is:
Average Force = -1.2 kg m/s / 0.0015 s = -800 N
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Given:
The mass of the hammerhead is m = 0.15 kg
The speed of the hammerhead is v = 8 m/s
The time for which the hammerhead is in contact with the nail is
[tex]\Delta t\text{ =}0.0015\text{ s}[/tex]To find the change in momentum
The impulse and the average force.
Explanation:
The change in momentum will be
[tex]\begin{gathered} \Delta p\text{ = mv} \\ =\text{ 0.15}\times8 \\ =1.2kg\text{ m/s} \end{gathered}[/tex]The impulse will be
[tex]\begin{gathered} Im\text{pulse = change in momentum} \\ =1.2kg\text{ m/s} \end{gathered}[/tex]The average force will be
[tex]\begin{gathered} F_{av}=\frac{Impulse}{\Delta t} \\ =\frac{1.2}{0.0015} \\ =800\text{ N} \end{gathered}[/tex]Final Answer: The change in momentum is 1.2 kg m/s
The impulse is 1.2 kg m/s.
The average force is 800 N
