Question

-One 2 kg lump of clay traveling at 33 m/s overtakes a second 1 kg lump traveling at 5.3 m/s. After collision they are stuck together. To the nearest tenth of a m/s what is their common velocity?

-If the second lump in the above problem was moving to the left before the collision what would be the answer?

Answer 1

m₁ = mass of first lump = 2 kg

V_1i = initial speed of first lump = 33 m/s

m₂ = mass of second lump = 1 kg

V_2i = initial speed of second lump = 5.3 m/s

V = common speed after collision,

Using conservation of momentum for inelastic collision ::

m₁ V_1i + m₂ V_2i = (m₁ + m₂) V

2 x 33 + 1 x 5.3 = (2 + 1) V

V = 23.8 m/s

b)

m₁ = mass of first lump = 2 kg

V_1i = initial speed of first lump = 33 m/s   moving right

m₂ = mass of second lump = 1 kg

V_2i = initial speed of second lump = -5.3 m/s moving left

V = common speed after collision,

Using conservation of momentum for inelastic collision ::

m₁ V_1i + m₂ V_2i = (m₁ + m₂) V

2 x 33 + 1 (-5.3) = (2 + 1) V

V = 20.2 m/s