Question

Will mark the brainliest

Answer 1

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:

$J = \Delta p = p_{f} - p_{i}$

Where:                                                                

$p_{f}$: is the final momentum

$p_{i}$: is the initial momentum

The initial momentum is given by:

$p_{i} = m_{1}v_{1_{i}} + m_{2}v_{2_{i}}$

Where 1 is for the hammer and 2 is for the nail.

Since the hammer is moving down (in the negative direction):

$v_{1_{i}} = -10 m/s$

And because the nail is not moving:

$v_{2_{i}}= 0$                      

$p_{i} = m_{1}v_{1_{i}} = 0.25 kg*(-10 m/s) = -2.5 kg*m/s$

Now, the final momentum can be found taking into account that the hammer remains in contact with the nail during and after the blow:

$p_{f} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Since the hammer and the nail are moving in the negative direction:

$v_{1_{f}}$ = $v_{2_{f}}$ = -9.7 m/s

$p_{f} = -9.7 m/s(7 \cdot 10^{-3} kg + 0.25 kg) = -2.49 kg*m/s$

Finally, the impulse is:

$J = p_{f} - p_{i} = - 2.49 kg*m/s + 2.50 kg*m/s = 0.01 kg*m/s$

Therefore, the impulse transferred to the nail is 0.01 kg*m/s.

I hope it helps you!