![]() An object travels with a velocity 4m/s to the east. Momentum and Collisions - Home || Printable Version || Questions with LinksĪnswers to Questions: All || #1-5 || #6-36 || #37-56 || 57-72ĥ7.Impulse Momentum Exam 1 and Problem Solutionsġ. (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. Where the "-" indicates that the impulse was opposite the original direction of motion. (b) The impulse is the product of force and time. 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. ![]() s impulse acts upon it in the direction of motion for 5.0 seconds.Ī 4.0-kg object has a forward momentum of 20.A resistive force of 6.0 N then impedes its motion for 8.0 seconds. Determine the final velocity of the object. 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. A 60-unit impulse will change the momentum by 60 units, either increasing or decreasing it. If the impulse is in the direction of an object's motion, then it will increase the momentum. 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. Since this impulse is "resistive" in nature, it will decrease the object's momentum by 48 units. The question asks for the object's velocity after encountering these two impulses. Since momentum is the product of mass and velocity, the velocity can be easily determined. ![]() A 3.0-kg object is moving forward with a speed of 6.0 m/s. The object then encounters a force of 2.5 N for 8.0 seconds in the direction of its motion. The object then collides head-on with a wall and heads in the opposite direction with a speed of 5.0 m/s. Determine the impulse delivered by the wall to the object. 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. This is equivalent to an impulse of 20 units (N The object encounters a force of 2.5 N for 8.0 seconds. Since this impulse acts in the direction of motion, it changes the object's momentum from 18 units to 38 units. Upon rebounding, the object has a momentum of -15 units (kg A final impulse is encountered when colliding with a wall. The -15 is the product of mass (3 kg) and velocity (-5 m/s). The "-" sign is used since the object is now moving in the opposite direction as the original motion. The collision with the wall changed the object's momentum from +38 units to -15 units. Thus, the collision must have resulted in a 53-unit impulse since it altered the object's momentum by 53 units.Ħ0. ![]() A 46-gram tennis ball is launched from a 1.35-kg homemade cannon. Given: m ball = 46 g = 0.046 kg m cannon = 1.35 kg v cannon = -2.1 m/s If the cannon recoils with a speed of 2.1 m/s, determine the muzzle speed of the tennis ball. The ball is in the cannon and both objects are initially at rest. The total system momentum is initially 0. After the explosion, the total system momentum must also be 0. Thus, the cannon's backward momentum must be equal to the ball's forward momentum. A 2.0-kg box is attached by a string to a 5.0-kg box. ![]() A compressed spring is placed between them. The two boxes are initially at rest on a friction-free track. The string is cut and the spring applies an impulse to both boxes, setting them in motion. ![]()
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