Answer:
1a). 4 times. 1b) 4. 1c) 1/4.
2a) 5 times. 1b) 5 1c) 1/5
3a) 7/3 times 3b) 7/3 3c) 3/7
4a) 8/5 times 4b) 8/5 4c) 5/8
Step-by-step explanation:
1) Assuming no external forces acting during the collision, total momentum must be conserved, so the following general expression applies:
m₁*v₁₀ + m₂*v₂₀ = m₁*v₁f + m₂*v₂f (1)
If we assume that the collision is perfectly inelastic, this means that both masses stick together after the collision, so v₁f = v₂f.
If m₂ is initially at rest, ⇒ v₂₀ = 0.
Replacing in (1) we get the expression of vf as a function of v₁₀, as follows:
vf = v₁₀*(m₁/(m₁+m₂)
So, for the four cases we have the following:
1) initial mass = m
final mass = m+3m = 4 m
⇒final mass / initial mass = 4
vf = v₀* (m/4m) = v₀/4 ⇒v₀/vf = 4
So, the velocity of the system will decrease by a factor of 4. The new velocity will be vf= v₀/4.
2) Applying the same considerations, we get:
2a) final mass / initial mass = 5
2b) vf = v₀* (m/5m) = v₀/5 ⇒v₀/vf = 5
2c) vf = v₀/5
3) Applying the same considerations, we get:
3a) final mass / initial mass = 7/3
3b) vf = v₀* (3m/7m) =3/7* v₀ ⇒v₀/vf = 7/3
3c) vf = 3/7*v₀
4) Applying the same considerations, we get:
4a) final mass / initial mass = 8/5
4b) vf = v₀* (5m/8m) = 5/8*v₀ ⇒v₀/vf = 8/5
4c) vf = 5/8*v₀