Answer:
The impulse transferred to the nail is 0.01 kg*m/s.
Explanation:
The impulse (J) transferred to the nail can be found using the following equation:
[tex] J = \Delta p = p_{f} - p_{i} [/tex]
Where:
[tex]p_{f}[/tex]: is the final momentum
[tex]p_{i}[/tex]: is the initial momentum
The initial momentum is given by:
[tex] p_{i} = m_{1}v_{1_{i}} + m_{2}v_{2_{i}} [/tex]
Where 1 is for the hammer and 2 is for the nail.
Since the hammer is moving down (in the negative direction):
[tex]v_{1_{i}} = -10 m/s [/tex]
And because the nail is not moving:
[tex]v_{2_{i}}= 0[/tex]
[tex] p_{i} = m_{1}v_{1_{i}} = 0.25 kg*(-10 m/s) = -2.5 kg*m/s [/tex]
Now, the final momentum can be found taking into account that the hammer remains in contact with the nail during and after the blow:
[tex] p_{f} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}} [/tex]
Since the hammer and the nail are moving in the negative direction:
[tex]v_{1_{f}}[/tex] = [tex]v_{2_{f}}[/tex] = -9.7 m/s
[tex] p_{f} = -9.7 m/s(7 \cdot 10^{-3} kg + 0.25 kg) = -2.49 kg*m/s [/tex]
Finally, the impulse is:
[tex] J = p_{f} - p_{i} = - 2.49 kg*m/s + 2.50 kg*m/s = 0.01 kg*m/s [/tex]
Therefore, the impulse transferred to the nail is 0.01 kg*m/s.
I hope it helps you!
