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 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.
m/s

Explanation:

Step1: Apply the law of conservation of momentum

The law of conservation of momentum states that \(m_1u_1 + m_2u_2=m_1v_1 + m_2v_2\). Let \(m_1 = 22\space g=0.022\space kg\), \(u_1 = 40.0\space m/s\), \(m_2 = 300\space g = 0.3\space kg\), \(u_2=- 2.45\space m/s\) (negative because the target is moving in the opposite direction of the knife's initial motion), and \(v_2 = 0\space m/s\) (target stops). We need to find \(v_1\).
Substitute the values into the formula: \((0.022\times40)+(0.3\times(- 2.45))=(0.022\times v_1)+(0.3\times0)\)

Step2: Simplify the left - hand side of the equation

First, calculate \(0.022\times40 = 0.88\) and \(0.3\times(-2.45)=-0.735\). Then \(0.88-0.735 = 0.022v_1\)

Step3: Solve for \(v_1\)

\(0.145 = 0.022v_1\), so \(v_1=\frac{0.145}{0.022}\approx6.59\space m/s\)

Answer:

\(6.59\space m/s\)