Question

A solid block of mass m2 = 1.14 kg, at rest on a horizontal frictionless surface, is connected to a relaxed spring (with spring constant k = 125 N/m whose other end is fixed. Another solid block of mass m1 = 2.27 kg and speed v1 = 2.00 m/s collides with the 1.14 kg block. If the blocks stick together, what is their speed immediately after the collision?

Answer 1

Answer:

v = 1 m/s

Explanation:

from the principle of conservation of momentum, we have following relation

initial momentum = final momentum

$m_{1}v_{1}+m_{2}v_{2} = (m_{1}+m_{2})v^{2}$

where

m1 = 1.14 kg

v1 = 2.0 m/s

m2 = 1.14 kg

v2  = 0 m/s

putting all value in the above equation

$1.14 *2.0+ 0 =(1.14+1.14)v^{2}$

$v =\frac{1.14*2.0}{1.14+1.14}$

v = 1 m/s