2a) A 1.0 kg steel ball strikes a wall with a speed of 30 m/s at an angle of the

Question

2a) A 1.0 kg steel ball strikes a wall with a speed of 30 m/s at an angle of theta=37 with the normal to the ground. It bounces off with the same speed and angle, as shown in the figure below. If the ball is in contact with the wall for 0.20s, what is the magnitude of the average force exerted on the ball by the wall? 2b)a wheel rotating with a constant acceleration turns through 30 revolutions during a 5.0s time interval. Its angular velocity at the end of this interval in 30 rad/s. What is the angular acceleration of the wheel?

Explanation / Answer

First break speed into horizontal and vertical components: V_x = 5.61*cos(52.7) = 3.400 m/s V_y = 5.61*sin(52.7) = 4.463 m/s FAT=MAV Force * delta_time = mass * delta_velocity Force = mass * delta_velocity / delta_time mass = 4 kg; delta_time = .231s; delta_velocity requires more work. V_y stays constant, and V_x has the same magnitude, but opposite direction, so it becomes -3.400 m/s. The difference then is V_xfinal - V_xinitial = -3.400 - 3.400 = -6.800 m/s Plug all that in: Force = 4 kg * -6.800 m/s / .231 s = -117.8 kg*m/s/s = -117.8 N

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