**Problem Description:**A cart of mass 4 kg is on a frictionless surface and compressed against an ideal spring with a spring constant of 192 N/m for a distance of 1.04 m. After it is released, it hits a second cart of mass 8 kg and bounces back while the second cart continues traveling onto a ramp where it reaches a height of \( h_2 = 0.67 \, \text{m} \). Find the velocity of the second cart in m/s immediately AFTER being hit by the first cart. Use \( g = 9.8 \, \text{m/s}^2 \) and retain your answer to two decimal places.**Diagrams Explanation:**The first diagram shows a cart (\( m_1 \)) of mass 4 kg compressed against a spring with spring constant \( k = 192 \, \text{N/m} \). The spring is on a horizontal frictionless surface. The second diagram shows the same cart (\( m_1 \)) traveling towards a second cart (\( m_2 \)) of mass 8 kg on the horizontal surface after being released from the spring. The second cart (\( m_2 \)) then moves up a ramp, reaching a height (\( h_2 \)) of 0.67 m. The task is to calculate the velocity of the second cart (\( m_2 \)) immediately after being hit by \( m_1 \).

Given data, Mass m1=4 kg Mass m2=8 kg Height h2=0.67 m Compression x=1.04 m Acceleration due to gravity g=9.8 m/s2Initial speed of m2=0 First calculating the speed of m2 after collision, v22=u22+2gh Substituting the values, we get, v22=u22+2ghv22=02+2×9.8×0.67v2=2×9.8×0.67=3.62 m/s Now, calculating the initial speed of m1, Using the law of conservation of energy, So, 12kx2=12m1u12u12=kx2m1u1=kx2m1u1=xkm1 Substituting the values, we get, u1=1.04×1924=1.04×48=1.04×6.93=7.21 m/sNow, Applying the law of conservation of momentum, m1u1=m1v1+m2v2v1=m1u1-m2v2m1 Substituting all the known values, we get, v1=4×7.21-8×3.624=28.84-28.964=-0.124=-0.03 m/s hence the velocity of the second cart is 3.62 m/s.

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A cart of mass 4 kg is on frictionless surface and compressed against an ideal spring with spring constant 192 N/m for a distance of 1.04 m. After it is released, it hits a second cart of mass 8 kg and bounces back while the second car continues traveling onto a ramp where it reaches a height of h₂ = 0.67 m. Find the velocity of the second cart in m/s immediately AFTER being hit by the first cart. Use g = 9.8 m/s² and retain your answer to two decimal places.

A cart of mass 4 kg is on frictionless surface and compressed against an ideal spring with spring constant 192 N/m for a distance of 1.04 m. After it is released, it hits a second cart of mass 8 kg and bounces back while the second car continues traveling onto a ramp where it reaches a height of h₂ = 0.67 m. Find the velocity of the second cart in m/s immediately AFTER being hit by the first cart. Use g = 9.8 m/s² and retain your answer to two decimal places.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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