Question

A boat with a mass of 1000 kg drifts with the current down a straight section of river parallel to the +x+x axis with a speed of 2.0 m/s. At t=0t=0 the boat engages its engines, and accelerates at 2.7 \,\mathrm{m/s^2}2.7m/s2. If the boat is oriented with its nose 45^\circ45∘ from the +x+x axis, which of the following is a possible value for the boat's momentum at t = 3.0\, \mathrm{s}t=3.0s after the engine had started?
Pick the correct answer

a. \vec{p} = (7700 \hat{i} + 5700 \hat{j}) \; \mathrm{kg \, m/s}p​=(7700i^+5700j^​)kgm/s
b. \vec{p} = (9500 \hat{i} - 1400 \hat{j}) \; \mathrm{kg \, m/s}p​=(9500i^−1400j^​)kgm/s
c. \vec{p} = (5700 \hat{i} + 9500 \hat{j}) \; \mathrm{kg \, m/s}p​=(5700i^+9500j^​)kgm/s
d. \vec{p} = ( 1400 \hat{i} - 7700 \hat{j}) \; \mathrm{kg \, m/s}p​=(1400i^−7700j^​)kgm/s
e. None of these are correct

Answer 1

Answer:

A.

Explanation:

Our values are defined by,

$m=1000kg\\v = 2m/s\\a = 2.7 \angle 45\°$

We can express also in a vectorial way, then

$v=2\hat{i} \rightarrow$no values in $\hat{j}$

Acceleration as follow,

$a= 2.7cos45 \hat{i} + 2.7 sin45 \hat{j}$

We know that velocity is given by,

$V(t) = v_0+at \rightarrow v_0 = 2$

We need to calculate for t=3, then

$v(3) = 2\hat{i} +(2.7cos45 \hat{i} + 2.7 sin45 \hat{j})*3$

$v(3) = (2\hat{i}+3*2.7cos45 \hat{i})+(3*2.7 sin45 \hat{j})$

$v(3) = (2+3*2.7cos45)\hat{i}+(5.7 sin45 )\hat{j}$

$v(3) = 7.7\hat{i}+5.7\hat{j}$

Our mass is 1000Kg, so the momentum is

$P = mv$

$P= 1000(7.7\hat{i}+5.7\hat{j})$

$P=7700\hat{i}+5700\hat{j}$