Question

1. a 1.30 kg object is dropped from a height of 6.5 m. how far did the object fall when its momentum is 6.0 kgm/s?
2. an average net force of 16.0 n acts on a box for 3.33×10⁻³ causing it to accelerate from rest to 3.50 m/s. what is the mass of the object?
3. a 0.500 kg object is thrown vertically upward with an average applied force of 8.20 n by a student. the force is applied through a displacement of 1.50 m.

Explanation:

Step1:Use the impulse - momentum theorem

The impulse - momentum theorem is \(J = \Delta p\). Here, \(J = F_{avg}\Delta t\) and \(\Delta p=m(v - u)\). Given \(F_{avg}=16.0\space N\), \(u = 0\space m/s\), \(v = 3.50\space m/s\), and \(\Delta t=3.33\times10^{-3}\space s\).
First, from \(J=\Delta p\), we have \(F_{avg}\Delta t=m(v - u)\). Since \(u = 0\space m/s\), the equation simplifies to \(F_{avg}\Delta t=mv\).

Step2:Solve for the mass \(m\)

We can re - arrange the equation \(F_{avg}\Delta t=mv\) to solve for \(m\). So, \(m=\frac{F_{avg}\Delta t}{v}\).
Substitute \(F_{avg}=16.0\space N\), \(\Delta t = 3.33\times10^{-3}\space s\), and \(v = 3.50\space m/s\) into the formula:
\(m=\frac{16.0\times3.33\times10^{-3}}{3.50}\)

$$m=\frac{0.05328}{3.50}$$

$$m = 0.0152\space kg$$

Answer:

\(m = 0.0152\space kg\)