Question

a box has a momentum of 1.30 kg*m/s to the right. a 12.8 N force pushes it to the right for 0.218 s. what is the final momentum of the box. PLEASE HELP

Answer 1

Answer:

4.09 kgm/s

Explanation:

Concept tested: Second Newton's law of motion

So what does the second Newton's law of motion state?

• According to the second Newton's law of motion, the resultant force and the rate of change of momentum are directly proportional.

• That is; $F=\frac{(mVf-mVo)}{t}$, where, mVf is the final momentum and mVo is the initial momentum.

In this case;

• Force = 12.8 N
• Initial momentum, mVo = 1.3 kgm/s
• Time, t = 0.218 seconds

Therefore; replacing the values of the variables in the formula;

we get;

$12.8N=\frac{(mVf-1.3kgm/s)}{0.218s}$

$(12.8N)(0.218s)=(mVf-1.3kgm/s)$

$2.7904+1.3=mVf$

Solving for final momentum;

$mVf =4.0904 kgm/s$

      = 4.09 kgm/s

Therefore, the final momentum of the box is 4.09 kgm/s