Question

Notify Me Bookmark

An object with mass m1 interacts with a second object m2 through a mutual attractive force. At the instant shown in the diagram below, object m1 has a velocity v1 while m2 has a velocity v2. Show that the total momentum of the system p = m1v1 + m2v2 does not change with time.

An object with mass m1 interacts with a second object m2 through a mutual attractive force. At the instant shown in the diagram below, object m1 has a velocity v1 while m2 has a velocity v2. Show that the total momentum of the system p = m1v1 + m2v2 does not change with time.

Image text
An object with mass m 1 interacts with a second object m 2 through a mutual attractive force. At the instant shown in the diagram below, object m 1 has a velocity v 1 while m 2 has a velocity v 2 . Show that the total momentum of the system p = m 1 v 1 + m 2 v 2 does not change with time.

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

A hanging weight, with a mass of m1 = 0.360 kg, is attached by a string to a block with mass m2 = 0.850 kg as shown in the figure below. The string goes over a pulley with a mass of M = 0.350 kg. The pulley can be modeled as a hollow cylinder with an inner radius of R1 = 0.0200 m, and an outer radius of R2 = 0.0300 m; the mass of the spokes is negligible. As the weight falls, the block slides on the table, and the coefficient of kinetic friction between the block and the table is μk = 0.250. At the instant shown, the block is moving with a velocity of vi = 0.820 m/s toward the pulley. Assume that the pulley is free to spin without friction, that the string does not stretch and does not slip on the pulley, and that the mass of the string is negligible. (a) Using energy methods, find the speed of the block (in m/s ) after it has moved a distance of 0.700 m away from the initial position shown. m/s (b) What is the angular speed of the pulley (in rad/s) after the block has moved this distance? rad/s

Solution
3 0

An m1 = 2.0 kg steel ball is initially moving to the right at a velocity of v14 = +10 m/s. The ball strikes an m2 = 3.0 kg steel ball that is initially at rest. The collision is elastic. After the collision, the velocity of m1 is v1f = −2.0 m/s. (a) What is the velocity of m2 after the collision, i.e. v2f? (b) How much kinetic energy did the ball with mass m1 lose? Equation: (a) m1v1i + m2v2i = m1v1f + m2v2f (b) In general, kinetic energy, K = 12 mv2 (a) v2f = −6.0 m/s (b) KE lost = 54 J (a) v2f = +8.0 m/s (b) KE lost = 96 J (a) v2f = +4.0 m/s (b) KE lost = 96 J (a) v2f = +10 m/s (b) KE lost = 75 J (a) v2f = +7.0 m/s (b) KE lost = 84 J
Solution
0 0

A hanging weight, with a mass of m1 = 0.360 kg, is attached by a cord to a block with mass m2 = 0.850 kg as shown in the figure below. The cord goes over a pulley with a mass of M = 0.350 kg. The pulley can be modeled as a hollow cylinder with an inner radius of R1 = 0.0200 m, and an outer radius of R2 = 0.0300 m; the mass of the spokes is negligible. As the weight falls, the block slides on the table, and the coefficient of kinetic friction between the block and the table is μk = 0.250. At the instant shown, the block is moving with a velocity of vi = 0.820 m/s toward the pulley. Assume that the pulley is free to spin without friction, that the cord does not stretch and does not slip on the pulley, and that the mass of the cord is negligible. (a) Using energy methods, find the speed of the block (in m/s ) after it has moved a distance of 0.700 m away from the initial position shown. m/s (b) What is the angular speed of the pulley (in rad/s) after the block has moved this distance? rad/s
Solution
0 0