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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 40.0 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
Step1: <Convert masses to kg>
The mass of the target \(m_{t}=300\ g = 0.3\ kg\), the mass of the knife \(m_{k}=22.5\ g=0.0225\ kg\).
Step2: <Apply conservation of momentum>
The initial momentum \(p_{i}=m_{k}v_{k}+m_{t}v_{t}\), where \(v_{k}\) is the initial speed of the knife (\(v_{k} = 40.0\ m/s\)) and \(v_{t}\) is the initial speed of the target (\(v_{t}=2.45\ m/s\)). The final momentum \(p_{f}=m_{k}v_{k}'\) (since the target is stopped, \(v_{t}' = 0\)).
By conservation of momentum \(p_{i}=p_{f}\), so \(m_{k}v_{k}+m_{t}v_{t}=m_{k}v_{k}'\).
Step3: <Solve for \(v_{k}'\)>
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\(72.7\ m/s\)