Question

A 0.30 kg mass is sliding on a horizontal, frictionless air track with a speed of 4.0 m/s when it instantaneously hits and sticks to a 2.0 kg mass initially at rest on the track. The spring constant is 100 N/m . The other end of the spring is fixed.a. Determine the following for the 0.30 kg mass immediately before the impact:i. Its linear momentum ii. Its KEb. Determine the following for the combined masses immediately after the impact:i. The linear momentum in. The KEc. Besides the fact that the objects stick together, what other clue lets you know that this was an inelastic collision

Answer 1

Answer

given,

mass of block 1, m = 0.3 Kg

speed of block 1, v = 4 m/s

mass of second block,M = 2 Kg

initial speed of block = 0 m/s

spring constant, k = 100 N/m

a) for block 1

  linear momentum before collision

   P₁ = m v = 0.3 x 4 = 1.2 Kg.m/s

  Kinetic energy

   $KE_1 = \dfrac{1}{2}mv^2$

   $KE_1 = \dfrac{1}{2}\times 0.3\times 4^2$

   $KE_1 =2.4\ J$

b) After impact

  final velocity calculation

 using conservation of momentum

  m v  = (m + M )v_f

   0.3 x 4 = 2.3 x v_f

    v_f = 0.522 m/s

   Linear momentum

   P₂ = (m+M) v_f

   P₂ = 1.5 x 0.522

   P₂ = 0.783 kg.m/s

   Kinetic energy

   $KE_2= \dfrac{1}{2}(M+m)v^2$

   $KE_2= \dfrac{1}{2}\times 2.3\times 0.522^2$

   $KE_2=0.313\ J$