1. Two blocks are sliding along a frictionless track. Block A (mass 3.75 kg) is

Question

1. Two blocks are sliding along a frictionless track. Block A (mass 3.75 kg) is moving to the right at 2.70 m/s. Block B (mass 4.50 kg) is moving to the left at 3.24 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 or right?

2. What is the total momentum of the system after the collision?

Left or Right?

3. 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 or Right?

Explanation / Answer

Here,

u1 = 2.70 m/s

u2 = -3.24 m/s

m1 = 3.75 Kg

m2 = 4.50 Kg

for the total momentum of the system before collision = m1 * u1 + m2 * u2

total momentum of the system before collision = 3.75 * 2.70 - 3.24 * 4.50

total momentum of the system before collision = -4.455 Kg.m/s

the total momentum of the system before collision is 4.455 Kg.m/s LEFT

2)

as there is no net external force acting on the system of blocks

total momentum of the system after collision = total momentum of the system before collision

total momentum of the system after collision = -4.455 Kg.m/s

the total momentum of the system after collision is 4.455 Kg.m/s LEFT

3)

let the final velocity is v

(m1 + m2 ) * v = total momentum

(3.75 + 4.50) * v = 4.455

v = 0.54 m/s left

the velocity of blocks after the collision is 0.54 m/s to the left

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

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