Question

a knife thrower throws a knife toward a 300 g target that is sliding in her direction at a speed of 2.45 m/s on a horizontal frictionless surface. she throws a 22.5 g knife at the target with a speed of 40.5 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.
72.7
establish the sign convention you wish to use and then try to write a statement of conservation of momentum that will allow you to determine the speed of the knife after passing through the target. m/s

Explanation:

Step1: Convert masses to kilograms

The mass of the target \(m_{t}=300\ g = 0.3\ kg\), the mass of the knife \(m_{k}=22.5\ g=0.0225\ kg\). The initial velocity of the target \(u_{t}=2.45\ m/s\), the initial velocity of the knife \(u_{k}=40.5\ m/s\), and the final velocity of the target \(v_{t}=0\ m/s\). Let the final velocity of the knife be \(v_{k}\).

Step2: Apply the law of conservation of momentum

The law of conservation of momentum states that \(m_{k}u_{k}+m_{t}u_{t}=m_{k}v_{k}+m_{t}v_{t}\).
Substitute the known values into the equation: \((0.0225\times40.5)+(0.3\times2.45)=(0.0225\times v_{k})+(0.3\times0)\).
First, calculate the left - hand side:
\(0.0225\times40.5 = 0.91125\) and \(0.3\times2.45 = 0.735\).
So, \(0.91125 + 0.735=0.0225v_{k}\).
\(1.64625 = 0.0225v_{k}\).

Step3: Solve for \(v_{k}\)

\(v_{k}=\frac{1.64625}{0.0225}\)
\(v_{k}=73.2\ m/s\)

Answer:

\(73.2\ m/s\)