please answer the second and third questions. What is the impulse on a 4.0 kg pa

Question

please answer the second and third questions.

What is the impulse on a 4.0 kg particle that experiences the force described by the graph in the figure? 1. Fx (N) 2500 2000 1500 1000 6 9 1 15 t (ms)16.5 -1000 Consider an inelastic collision between a 1 Kg cart moving to the right at 2 m/s and a 2 Kg cart moving to the left at 4 m/s. What is the speed and direction of the 2 carts after they have stuck together? 2. A mass mi traveling to the right with a speed vi makes a glancing colision with a mass m2 initially at rest. After the collision the masses have speeds Vi and v2 and move in directions and 02, as shown below. Determine the velocity of m2 after collision (i.e. find v2) 3. vi 3 m/s mi 1 kg 1-5 m/s mi 30 m2 2 kg

Explanation / Answer

Ans 2 :

The momentum will remain conserved for inelastic collision

Thus Initiam momentum = final momentum

Here we have m1 = 1 Kg and v1 = +2m/s (right direction)

m2 = 2 Kg and v2 = -4m/s (left direction)

m1v2 + m2v2 = (m1+m2)Vfinal

1*2 + 2*(-4) = (1+2)*Vfinal

Vfinal = -2m/s (Left direction)

Ans 3:

Momentum will conserve in x direction and y direction

For x direction

Initial momentum = m1v1 = 1*5 = 5 Kgm/s

Final momentum = m1*v1'(cos(30)) + m2v2'(cos(theta))

5 = 3*(0.866) + 2v2'(cos(theta))

1.201 = v2'(cos(theta)) ----- (1)

For Y- direction :

0 = 3*sin(30) - 2*v2'*sin(theta)

v2'sin(theta) = 0.75 ---- (2)

Dividint 2 and1 we get

tan(theta) = 0.75/1.201 = 0.6245

theta = 31.98 degrees

from equation 2 we have

v2' = 0.75/sin(theta) = 0.75/0.529 = 1.42 m/s

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