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.

Explanation:

Step1: Apply conservation of momentum

The formula for conservation of momentum is \(m_1u_1 + m_2u_2=m_1v_1 + m_2v_2\). Let \(m_1 = 22.5\space g=0.0225\space kg\), \(u_1 = 40.0\space m/s\), \(m_2 = 300\space g = 0.3\space kg\), \(u_2=2.45\space m/s\), and \(v_2 = 0\space m/s\). We need to find \(v_1\).
Substitute the values into the formula: \((0.0225\times40)+(0.3\times2.45)=(0.0225\times v_1)+(0.3\times0)\)

Step2: Simplify the equation

First, calculate the left - hand side:
\(0.0225\times40=0.9\) and \(0.3\times2.45 = 0.735\)
So, \(0.9 + 0.735=0.0225v_1\)
\(1.635 = 0.0225v_1\)

Step3: Solve for \(v_1\)

\(v_1=\frac{1.635}{0.0225}\)

Answer:

\(v_1 = 72.7\space m/s\)