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?
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Ответ:
1.5kg
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
Ответ: