Question

please be descriptive so i actually understand it.

A 2.4 kg block slides along a frictionless surface at 1.3 in/s. A second block, sliding at a faster 4.2 m/s, collides with the first from behind and sticks to it. The final velocity of the combined blocks is 2.3 m/s. What was the mass of the second block? Express your answer to two significant figures and include the appropriate units.

Answer 1

This is a case of inelastic collision ,where only momentum is conserved always.[not energy]
So,
v1 = 1.3 m/s , v2 = 4.2 m/s
m1 = 2.4 kg, to find , m2= ??
given,
m3 =(m1 +m2) , v3 = 2.3 m/s

So, using, m1v1 +m2v2 = m3v3,
we get, 2.4 x 1.3 + m2 x 4.2 = (2.4+m2) x 2.3
3.12 + m2 x 4.2 = 5.52 + 2.3xm2

1.9 m2 = 2.4
So, m2 = 1.26 Kg --> Answer