Question

A 2.5 kg block slides along a frictionless surface at 1.5 m/s.A second block, sliding at a faster 4.1 m/s , collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.5 m/s. What was the mass of the second block?

Answer 1

1.5kg

Step-by-step explanation:

Given data

mass m1= 2.5kg

mass m2=??

velocity of mass one v1= 1.5m/s

velocity of mass two v2= 4.1m/s

common velocity after impact v= 2.5m/s

Let us apply the formula for the conservation of linear momentum for inelastic collision

The expression is given as

m1v1+ m2v2= v(m1+m2)

substitute

2.5*1.5+ m2*4.1= 2.5(2.5+m2)

3.75+4.1m2= 6.25+2.5m2

collect like terms

3.75-6.25= 2.5m2-4.1m2

-2.5= -1.6m2

divide both sides by -1.6

m2= -2.5/-1.6

m2= 1.5 kg

Hence the second mass is 1.5kg