Mass and Kinetic Energy
Example problems involving collisions 1) On a horizontal frictionless surface a puck of mass m initially at speed u collides head-on (without rotation) with a stationary puck of mass M. Find the velocities of both puck after the collision if: i) the collision is fully elastic ii) the collision if fully inelastic. i) momentum: kinetic energy: mu = mv+MV (+ve in direction of initial u) 1 /2 m u2 = 1/2 m v2 + 1/2 M V2
2 eqns in 2 unknowns: V = (u - v) m/M substitute in K eqn: u2 = v2 + (M/m) V2 = v2 + (M/m) (u - v)2 (m/M)2 = v2 + (u - v)2 (m/M) let ρ = (m/M) ⇒ v2 (1 + ρ) - 2ρ u v + u2 (ρ - 1) = 0 quadratic eqn: b2-4ac = 4ρ2 u2 - 4 (1 + ρ) (ρ - 1) u2 = 4ρ2 u2 - 4 (ρ2 - 1) u2 = 4u2
v = [2ρ u ± (4 u2)1/2]/{2 (1 + ρ)} = [2ρ u ± 2 u]/{2 (1 + ρ)} = u (ρ ± 1)/(1 + ρ) + ⇒ v = u , and V = 0 (no collision occurs!) - ⇒ v = u (ρ - 1)/(1 + ρ) , and V = ρ (u - v) = 2ρ2 u/(1 + ρ) e.g. ρ = 1 ⇒ v = 0, V = u . as ρ → 0 ⇒ v → - u , V → 0
ii) momentum: let ρ = (m/M) kinetic energy: ratio: then
m u = (m + M) v v = u 1/(1 + 1/ρ)
1
⇒
v = u m/(m + M)
before: K1 =
/2 m u2 ,
after:
K2 =
1
/2 (m + M) v2
K2/K1 = (1 + 1/ρ) v2/u2 = 1/(1 + 1/ρ) ⇒ v = u/2 , K2/K1 = 1/2 . as ρ → 0 ⇒ v → 0 , K2/K1 → 0
e.g. ρ = 1
2) A point mass m swings under gravity from a fixed pivot on a massless cord through an angle θ to collide with a stationary block of mass M. Assuming a fully elastic collision find the distance the block will slide if the dynamic friction is µ. during swing down m acquires speed u: m g Δh = 1/2 m u2 ⇒ u = [2g R(1 - cosθ)]1/2 let ρ = (m/M) then speed of block just after impact is: V = 2ρ2 u/(1 + ρ) v = u (ρ - 1)/(1 + ρ)
so after collision, block M has KE:
K = 1/2 M V2
friction force does work on the sliding block to bring its speed to zero: W = f Δx = µ M g Δx = K ⇒ Δx = K/(µ M g) =
1
/2 V 2/(µ g)
Δx = 2ρ4 u2/[(1 + ρ)2(µ g)] = 4ρ4 [R(1 - cosθ)]/[(1 + ρ)2µ]
e.g. for m= 1kg, M = 4 kg, R = 1 m, µ = 0.25, θ = 60o: ⇒ ρ = 1/4 , u =
2 eqns in 2 unknowns: V = (u - v) m/M substitute in K eqn: u2 = v2 + (M/m) V2 = v2 + (M/m) (u - v)2 (m/M)2 = v2 + (u - v)2 (m/M) let ρ = (m/M) ⇒ v2 (1 + ρ) - 2ρ u v + u2 (ρ - 1) = 0 quadratic eqn: b2-4ac = 4ρ2 u2 - 4 (1 + ρ) (ρ - 1) u2 = 4ρ2 u2 - 4 (ρ2 - 1) u2 = 4u2
v = [2ρ u ± (4 u2)1/2]/{2 (1 + ρ)} = [2ρ u ± 2 u]/{2 (1 + ρ)} = u (ρ ± 1)/(1 + ρ) + ⇒ v = u , and V = 0 (no collision occurs!) - ⇒ v = u (ρ - 1)/(1 + ρ) , and V = ρ (u - v) = 2ρ2 u/(1 + ρ) e.g. ρ = 1 ⇒ v = 0, V = u . as ρ → 0 ⇒ v → - u , V → 0
ii) momentum: let ρ = (m/M) kinetic energy: ratio: then
m u = (m + M) v v = u 1/(1 + 1/ρ)
1
⇒
v = u m/(m + M)
before: K1 =
/2 m u2 ,
after:
K2 =
1
/2 (m + M) v2
K2/K1 = (1 + 1/ρ) v2/u2 = 1/(1 + 1/ρ) ⇒ v = u/2 , K2/K1 = 1/2 . as ρ → 0 ⇒ v → 0 , K2/K1 → 0
e.g. ρ = 1
2) A point mass m swings under gravity from a fixed pivot on a massless cord through an angle θ to collide with a stationary block of mass M. Assuming a fully elastic collision find the distance the block will slide if the dynamic friction is µ. during swing down m acquires speed u: m g Δh = 1/2 m u2 ⇒ u = [2g R(1 - cosθ)]1/2 let ρ = (m/M) then speed of block just after impact is: V = 2ρ2 u/(1 + ρ) v = u (ρ - 1)/(1 + ρ)
so after collision, block M has KE:
K = 1/2 M V2
friction force does work on the sliding block to bring its speed to zero: W = f Δx = µ M g Δx = K ⇒ Δx = K/(µ M g) =
1
/2 V 2/(µ g)
Δx = 2ρ4 u2/[(1 + ρ)2(µ g)] = 4ρ4 [R(1 - cosθ)]/[(1 + ρ)2µ]
e.g. for m= 1kg, M = 4 kg, R = 1 m, µ = 0.25, θ = 60o: ⇒ ρ = 1/4 , u =