1. A 3.00 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 6

Question

1. A 3.00 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.200 seconds, what is the average force exerted by the wall? 60?" 600 2. A 12.0 gram wad of sticky clay is hurled horizontally at a 100 gram wooded block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.5m before coming to rest. If the coefficient of friction between the block and the surface is 0.650, what was the speed of the clay immediately before impact? . A 10.0 kg block is released from point A in the figure. The track is frictionless except 6 for the portion between points B and C. which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2,250 N/m and compresses the spring 0.300 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between B and C 00m . A 90.0 kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0 kg opponent running north with a speed of 5.00 m/s is tackled by a 95.0 kg opponent running north with a speed of 3.00 m/s. If the collision is perfectly inelastic, Calculate the speed and direction of the players just after the tackle Determine the mechanical energy lost as a result of the collision. a. b. 5. A skier of mass 70.0 kg is pulled up a slope by a motor-driven cable a. How much work is required to pull him a distance of 60.0 m up a 30.0 slope (assumed frictionless) at a constant speed of 2.00 m/s? b. A motor of what power is required to perform the task?

Explanation / Answer

1) Impulse in x-direction = change in momentum in x-direction

F_avgx*delta_t = m*(v2x - v1x)

F_avgx = m*(v2x - v1x)/delta_t

= 3*(-10 - 10)*sin(60)/0.2

= -260 N

negative sign indicates the force acting towards -x axis.

2)
let m = 12 grams
M = 100 grams

let v is the speed of the block just after the impact.

acceleration of block, a = -g*mue_k

= -9.8*0.65

=    -6.37 m/s^2

now use, vf^2 - vi^2 = 2*a*d

0^2 - v^2 = 2*(-6.37)*7.5

==> v = sqrt(2*6.37*7.5)

= 9.77 m/s

let u is the speed of the clay just before the impact

apply conservation of momentum

m*u = (m + M)*v

u = (m + M)*u/m

= (12 + 100)*9.77/12

= 91.2 m/s

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