A knife thrower throws a knife toward a 300 g target that is sliding in her direction at a speed of 2.30 m/s on a horizontal frictionless surface. She throws a 22.5 g knife at the target with a speed of 40.0 m/s. The target is stopped by the impact and the knife passes through the target. Determine the speed of the knife (in m/s) after passing through the target.

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whitneytr12 whitneytr12 · Answer 1 · 2021-04-17T03:07:56+02:00

Answer:

The speed of the knife after passing through the target is 9.33 m/s.

Explanation:

We can find the speed of the knife after the impact by conservation of linear momentum:

[tex] p_{i} = p_{f} [/tex]

[tex] m_{k}v_{i_{k}} + m_{t}v_{i_{t}} = m_{k}v_{f_{k}} + m_{t}v_{f_{t}} [/tex]

Where:

[tex] m_{k}[/tex]: is the mass of the knife = 22.5 g = 0.0225 kg

[tex] m_{t}[/tex]: is the mass of the target = 300 g = 0.300 kg

[tex] v_{i_{k}}[/tex]: is the initial speed of the knife = 40.0 m/s

[tex] v_{i_{t}} [/tex]: is the initial speed of the target = 2.30 m/s

[tex]v_{f_{k}}[/tex]: is the final speed of the knife =?

[tex] v_{f_{t}} [/tex]: is the final speed of the target = 0 (it is stopped)

Taking as a positive direction the direction of the knife movement, we have:

[tex] m_{k}v_{i_{k}} - m_{t}v_{i_{t}} = m_{k}v_{f_{k}} [/tex]

[tex] v_{f_{k}} = \frac{m_{k}v_{i_{k}} - m_{t}v_{i_{t}}}{m_{k}} = \frac{0.0225 kg*40.0 m/s - 0.300 kg*2.30 m/s}{0.0225 kg} = 9.33 m/s [/tex]

Therefore, the speed of the knife after passing through the target is 9.33 m/s.

I hope it helps you!