Two blocks are sliding along a frictionless track. Block A (mass 4.99 kg) is mov

Question

Two blocks are sliding along a frictionless track. Block A (mass 4.99 kg) is moving to the right at 1.70 m/s. Block B (mass 5.99 kg) is moving to the left at 2.04 m/s. Assume the system to be both Block A and Block B. What is the total momentum of the system before the collision? Left Right Submit Answer Tries 0/10 What is the total momentum of the system after the collision? Left Right Submit Answer Tries 0/10 The two blocks stick together and move off together. What is the magnitude and direction of the velocity of the blocks after the collision? Left Right

Explanation / Answer

here,

mA = 4.99 kg at v1 = 1.7 m/s

mB = 5.99 kg at v2 = - 2.04 m/s

the total momentum of the system before the collision , Pi = mA*v1 + mB*v2

Pi = 4.99 * 1.7 - 5.99 * 2.04

Pi = -3.74 kg.m/s

the total momentum of the system before the collision is 3.74 towards left

using conservation of momentum

Pi = Pf

the total momentum of the system after the collision is 3.74 kg.m/s towards left

let the final velocity be v

Pf = (mA + mB) * v

3.74 = ( 4.99 + 5.99) * v

v = 0.34 m/s

the final velocity is 0.34m/s towards left

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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