Question

for both questions

Two objects with masses represented by m, and m2 are moving such that their combined total momentum has a magnitude of 17.7 kg. m/s and points in a direction 71.5° above the positive x-axis. Object my is moving in the x direction with a speed of V1 = 2.90 m/s and my is moving in the y direction with a speed of v2 = 3.07 m/s. Determine the mass of each object in kilograms. kg m2 - kg 19. (-/3 Points) DETAILS SERCPWA11 6.WA.008. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER A particle with a mass of 2.15 kg is acted on by a force Fx acting in the x- direction. If the magnitude of the force varies in time as shown in the figure below, determine the following. F.(N) 8 6 4 2 t(s) 2 3 4 (a) impulse of the force (in kg - m/s) kg.m/s (b) final velocity of the particle (in m/s) if it is initially at rest m/s (c) Find the final velocity of the particle (in m/s) if it is initially moving along the x-axis with a velocity of -2.20 m/s. m/s

Answer 1

Solution of Q-1:

Let P is the magnitude ot total momentum and it make an angle $\theta$ from positive X- axis, then

the X - component of total momentum is given by

$P\cos \theta =m_{1}v_{1x}+m_{2}v_{2x}$ (1)

and the Y - component of total momentum is given by

(2)

P sin = mjuly + m2U2y

From the question:

$P=17.7kgm/s\\\\ \theta=71.5^{\circ}\\\\ v_{1x}=2.9m/s\\\\ v_{1y}=0.0m/s\\\\ v_{2x}=0.0m/s\\\\ v_{2y}=3.07m/s\\\\$

Putting these valuese in eqⁿ (1), we get

$17.7\times \cos 71.5 =2.9\times m_{1}+0\times m_{2}\\\\ \Rightarrow m_{1}=\frac{17.7\times \cos 71.5}{2.9}\\\\ \Rightarrow m_{1}=1.94kg\\\\$ (3)

Similarly, putting these valuese in eqⁿ (2), we find

$17.7\times \sin 71.5 =0\times m_{1}+3.07\times m_{2}\\\\ \Rightarrow m_{2}=\frac{17.7\times \sin 71.5}{3.07}\\\\ \Rightarrow m_{2}=5.47kg\\\\$ (4)