The figure below shows two block initially at rest which split apart due to the release of a spring or explosion. The mass of block 1 is 9.7 kg and the mass of block 2 is 2.9 kg and the after collision velocity of block 1 is 5.9 m/s to the left (minus direction). Determine the after collision velocity of block 2.
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The figure below shows two block initially at rest which split apart due to the release of a spring or explosion. The mass of block 1 is 9.7 kg and the mass of block 2 is 2.9 kg and the after collision velocity of block 1 is 5.9 m/s to the left (minus direction). Determine the after collision velocity of block 2.
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- Needs Complete typed solution with 100 % accuracy.(a) Calculate the x and y components of ball 1’s momentum before and after the collision. The mass of each pool ball is 0.16 kg. Remember that momentum is a vector quantity and the signs of the components indicate the direction of travel. (b) Use a before and after analysis to calculate the x and y components of ball 2’s velocity after the collision.A block is released from rest at the top of the frictionless ramp shown in the figure. Mass 1 is 1.27 kg and it is released from a height of 1.56 m. The block collides elastically with block 2, initially at rest. Mass 2 is 4.12 kg. How far back up the ramp does mass 1 move after the collision? (Take the initial direction of block one as positive.)
- A 0.230 kg billiard ball that is moving at 3.70 m/s strikes the bumper of a pool table and bounces straight back at 2.96 m/s (80% of its original speed). The collision lasts 0.0200 s. (Assume that the ball moves in the positive direction initially.) (a)Calculate the average force (in N) exerted on the ball by the bumper. (Indicate the direction with the sign of your answer.) Answer _____________ N (b)How much kinetic energy in joules is lost during the collision? (Enter the magnitude.) Answer_______________J (c)What percent of the original energy is left? Answer ________________ %Needs Complete typed solution with 100 % accuracy.Lucy is cruising through space in her new spaceship. As she coasts along, a tiny spacebug drifts into her path and bounces off the window. Consider several statements concerning this scenario. Evaluate each statement according to the law of momentum conservation and match it to the appropriate category. Must be true according to the law of momentum conservation Must be false according to the law of momentum conservation Answer Bank Not determined by the law of momentum conservation If the spacebug had stuck to the spaceship instead of bouncing off, momentum would not have been conserved for this interaction. The change in the spacebug's momentum during the collision is greater than the change in the spaceship's momentum. The total chanaa in momentum for this interaction is nem The total momentum of the spaceship and spacebug before the collision is equal to their total momentum afterward. 日示全部 X
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- Can you help me with this problem A 0.156 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.900 m/sm/s. It has a head-on collision with a 0.291 kg glider that is moving to the left with a speed of 2.15 m/s Suppose the collision is elastic.Find the magnitude of the final velocity of the 0.156 kg glider in m/s. Find the magnitude of the final velocity of the 0.291 kg glider in m/sUse the worked example above to help you solve this problem. A car with mass 1.54 x 103 kg traveling east at a speed of 29.7 m/s collides at an intersection with a 2.47 x 103 kg van traveling north a speed of 17.2 m/s, as shown in the figure. Find the magnitude and direction of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (that is, they stick together) and assuming that friction between the vehicles and the road can be neglected. 0.8 magnitude Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s direction 47.11 ° counterclockwise from the +x-axismoving speeds of v₁ = 5.2 m/s m₂ = 2m₁. After the collision Two Particles are and V₂0 and mass mi Of 90° relative to its direction prior to elastic. The collision is is observed to be moving at an angle the Collision, Define a Suitable coordinate system and determine the final velocities (including direction) of each Partille, It also may be useful to remember that cos(20) = Cos²-sin².