Question

Can someone help with this one question please

Answer 1

Answer:

v₃ = 1.25[m/s]

Explanation:

In order to solve this problem, we must use the principle of conservation of the amount of movement and momentum.

That is, the amount of movement is preserved before and after the collision. Let's propose an equation, in which the elements to the left of the equal sign represent the moment before the collision and to the right the moment after the collision.

$P=m*v$

P = lineal momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

$(m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}$

m₁ = mass of the first railroad = 2096 [kg]

m₂ = mass of the second railroad = 6280 [kg]

v₁ = velocity of the first railroad before the collision = 5 [m/s]

v₂ = velocity of the second railroad before the collision = 0 (initially at rest)

v₃ = velocity of the group after the collision [m/s]

Now replacing:

$(2096*5)+(6280*0)=(2096+6280)*v_{3}\\10480=8376*v_{3}\\v_{3}=1.25[m/s]$