Background In a perfectly inelastic collision, the colliding objects stick toget

Question

Background In a perfectly inelastic collision, the colliding objects stick together or become entangled after the collision. Thus, conservation of momentum gives the following equation. m1v1i + m2v2i = (m1 + m2)vf Solve this for the final velocity vf: vf = ( (m1v1i + m2v2i) / (m1 + m2)) (1) Explore First, adjust the sliders so that m1 = m2 = 13 kg. Then set v1i = 4 m/s, and v2i = -4 m/s. Before you click start, let's use Equation 1 to make a prediction: vf = m/s Now click start to see if your prediction agrees with the animation. Next, predict what will happen if m1 = m2 = 4 kg, v1i = 12 m/s and v2i = -12 m/s. vf = m/s Display in a New Window Now click start to see if you prediction agrees with the animation. Sometimes, it pays to be careful with equations such as Equation (1). Consider one more case: m1 = m2 = 4 kg, v1i = +12 m/s and v2i = +12 m/s. This the same as the above case except that both velocities are positive. Take a minute to make a prediction, then click start to see if you were right. Remarks Did you predict 12 m/s? Obviously, this cannot be correct (play the animation to verify). The only safeguard that comes with Equation (1) is your own intuition! Exercise 8.8 Suppose m1 = 8.99 kg, m2 = 9.92 kg and vf = 5.02 m/s. If v1i = 10.3 m/s, what was v2i? v2i = m/s

Explanation / Answer

conservation of momentum

m1v1i + m2v2i = (m1 + m2)vf

given values

m1 = 8.99 kg, m2 = 9.92 kg, v1i = 10.3 m/s, vf = 5.02 m/s

8.99*10.3 + 9.92*v2i = (8.99 + 9.92)*5.02

v2i = 0.235 m/s

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.