1.)Two hockey players strike a puck of mass 0.159 kg with their sticks simultane
ID: 2021536 • Letter: 1
Question
1.)Two hockey players strike a puck of mass 0.159 kg with their sticks simultaneously, exerting forces of 1.10 103 N, directed west, and 1.05 103 N, directed 30.0° east of north. Find the instantaneous acceleration of the puck.magnitude _________
direction_________ north of west
2.)A golf ball of mass 4.50 10-2 kg is struck by a club. Contact lasts 1.59 10-4 s, and the ball leaves the tee with a horizontal velocity of 55.0 m/s. Compute the average force the club exerts on the ball by finding its average acceleration.
magnitude
direction
3.)A 114 kg fullback runs at the line of scrimmage.
(a) Find the constant force that must be exerted on him to bring him to rest in a distance of 1.5 m in a time interval of 0.30 s.
___________________opposite the fullback's direction of motion
(b) How fast was he running initially?
4.)A small weight hangs from a string attached to the rearview mirror of a car accelerating at the rate of 1.17 m/s2. What angle does the string make with the vertical?
Explanation / Answer
1) We want to use F = ma unfortunately Newtons and north both use N, so call north 'up'. sideways force is 1.10 x 10^3N W + (1/2) 1.05 x 10^3N E The 1/2 is since the component east in a force 30 deg east of north is 1/2 (sin30 = 1/2) 1.10 x 10^3N - (1/2) 1.05 x 10^3N = 575N W The up and down force comes from the second player the component up is 1050N*cos30 = 909.3N Add these two forces like sides of a right triangle to get total force Ftot = SQRT(575N^2 + 909.3N^2) = 1075.9N We should have seen this; the east west struggle resulted in about half the total force west and the second guy had Sqrt(3)/2 left going up. These are the two legs of a 30, 60, 90 right triangle with the hypotenuse equal to the starting force. Instantaneous acceleration = Ftot/m = 1075.9N/0.159kg = 6,776 m/s^2 and the direction is 57.7 deg north of west. 2) Time of acceleration is 1.59 x 10 ^-4 s starting velocity is zero final velocity is 55m/s Let's use Vf = Vo + at to find average a a = Vf/t = (55m/s)/1.59 x 10^-4 s = 3.69E+05m/s^2 now let's use this to find average force F = ma F = 4.50 x 10 -2 kg*3.69E+05m/s^2 = 1.66E+04N or 16,600 N Direction horizontal. We are not told an angle, so assume all of the force is directed horizontally. I htink time may be up, so I will stop here.
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