A 5.0 kg mass with an initial velocity of 4.0 m/s due cast collides with a 4.0 k

Question

A 5.0 kg mass with an initial velocity of 4.0 m/s due cast collides with a 4.0 kg mass with an initial velocity of 2.0 m/s due west. After the collision the 5.0 kg mass has a velocity of 1.2 m/s due south. What is the magnitude and direction of the velocity of the 4.0 kg mass after the collision? A cooling fan is turned off when it is running at 58000 rev/hr. it turns 26000 revolutions before coming to a rest. a. What was the fan's annular acceleration? b. How long did it take it to come to s complete stop? An ideal spring with spring constant 9 N/m is sitting vertically on the ground an equilibrium length of 65 cm. A 10 kg mass is placed on top of the spring and compresses it. Calculate: a. The displacement of the spring in metres; b. The potential energy of the spring in joules; c. If all the potential energy of the spring were transferred to the mass what initial velocity would it receive in m/s? d. How high would it travel in the air in metres if all the potential energy were converted to kinetic energy? Two masses m_1 (15kg) and m_2 (25 kg) are shown in the figure below. Mass 1 sits on top of the table; mass 2 is hanging from a massless, inextensible tether over a non-rotating, friction less pulley. The coefficient of static friction is given as mu_s, and coefficient of kinetic friction is given as mu_k a. If the blocks are initially in equilibrium, determine the numerical value of mu_s. b. In the ease where block 1 is accelerating to the right at 2.5 m/s^2, determine the numerical value of mu_s. c. For the same scenario in part (C) determine the numerical value of T. A 12 kg box starts from rest and slides down a plane with a coefficient of kinetic friction of 0.23 and constant velocity of 6.5 m/s. Determine the angle of the incline?

Explanation / Answer

8)

let,

mass m1=5kg, u1=4 m/sec and v1=1.2 m/sec

mass m2=4kg, u2=2 m/sec and final velocity is v2

by using law of conservation,

along x axis,

m1*u1x+m2*u2x=m1*v1x+m2*v2x

5*4+4*(-2)=0+4*v2x

===> v2x=3 m/sec

and

along y axis,

m1*u1y+m2*u2y=m1*v1y+m2*v2y

0=5*1.2+4*v2y

===> v2y=-1.5 m/sec

magnitude of v2=sqrt(vx^2 +vy^2)

=sqrt(3^2 +(-1.5)^2)

=3.27 m/sec

and

dierction, tan(theta)=v2y/v2x

=-1.5/3

===> theta=26.56 degrees w r to -x axis

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