![]() ![]() Here the object begins with a momentum of 18 units (kg Like the previous problem, this problem is best solved by thinking through it conceptually using the impulse-momentum change principle. Determine the impulse delivered by the wall to the object. The object then collides head-on with a wall and heads in the opposite direction with a speed of 5.0 m/s. The object then encounters a force of 2.5 N for 8.0 seconds in the direction of its motion. A 3.0-kg object is moving forward with a speed of 6.0 m/s. Since momentum is the product of mass and velocity, the velocity can be easily determined. The question asks for the object's velocity after encountering these two impulses. Since this impulse is "resistive" in nature, it will decrease the object's momentum by 48 units. This is equivalent to an impulse of 48 units (N The object then encounters a resistive force of 6.0 N for 8.0 s. If the impulse is in the direction of an object's motion, then it will increase the momentum. A 60-unit impulse will change the momentum by 60 units, either increasing or decreasing it. It then encounters an impulse of 60 units (N This question is best thought about conceptually using the principle that an objects momentum is changed when it encounters an impulse and the amount of change in momentum is equal to the impulse which it encounters. Determine the final velocity of the object. A resistive force of 6.0 N then impedes its motion for 8.0 seconds. s impulse acts upon it in the direction of motion for 5.0 seconds.A 4.0-kg object has a forward momentum of 20. ![]() So if impulse is known and time is known, force can be easily determined.į = Impulse/t = (-16.7 N s) / (0.100 s) = -167 Nĥ8. (b) The impulse is the product of force and time. Where the "-" indicates that the impulse was opposite the original direction of motion. (a) determine the impulse with the wall, (b) determine the force of the wall on the ball.Īnswer: Answer: (a) -16.7 N s (b) -167 N A 0.530-kg basketball hits a wall head-on with a forward speed of 18.0 m/s. Momentum and Collisions - Home || Printable Version || Questions with LinksĪnswers to Questions: All || #1-5 || #6-36 || #37-56 || 57-72ĥ7. It also replaces the need to build natural gas peak power plants used only during high demand.The Review Session » Momentum and Collisions » Answers Q#57-72 Momentum and Collisions Review This process was designed to level the load of nearby nuclear power plants on the grid. Maximum water flow is over 33 million US gallons (120, 000 m 3) per minute. Electrical generation can begin within two minutes with peak electric output of 1872 MW achieved in under 30 minutes. During periods of peak demand water is released to generate power. At night, during low demand for electricity, the turbines run in reverse to pump water 363 feet (111 m) uphill from Lake Michigan into the reservoir. The power plant consists of six reversible turbines that can each generate 312 megawatts of electricity for a total output of 1, 872 megawatts. 4 km 2) reservoir is located on the banks of Lake Michigan. 6 km) wide which holds 27 billion US gallons (100 Gl) of water. 03 -05 ENERGY IN HUMANS AND THE WORLD Ludington Pumped Storage Power Plant It consists of a reservoir 110 feet (34 m) deep, 2.
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