Did I Solve the Inelastic Collision Pendulum Problem Correctly?

[ss85]

Alright, I did all the work for this problem, but do not know the correct answers and therefore cannot check my work with 100% confidence. So if somebody could look it over and let me know if I did everything correctly or not, I'd appreciate it. The answer I calculated for part e seemed kind of high to me.

Homework Statement

Two small spheres made of Play Doh hang from massless strings of length 2 meters. Sphere m₁ = 1kg is pulled to the left to an angle of θ₀ = 60° and has zero initial speed. It collides inelastically with sphere m₂ = 2kg, which is initially at rest. After the collision the two masses stick together and they act as a single pendulum.

(a) What is the speed of m₁ just before the collision?
(b) What is the speed of the two masses right after the collision?
(c) What is the ratio of mechanical energy before and after the collision?
(d) What is the amplitude θ_max of the swinging pendulum?
(e) Assume that the collision lasted 10^-3 seconds. What was the average force that the mass m₁ was acting on mass m₂ during the collision?

Homework Equations

KPE = 1/2 m v²
GPE = mgh
P = mv
F_avg = I/(Δt)

The Attempt at a Solution

(a) v = 4.43 m/s
mgh = 1/2 mv²
9.8 * 1 = 1/2 v²
v² = 19.6
v = 4.43 m/s

(b) v₂ = 1.48 m/s
m₁v₁ = (m₁ + m₂) v₂
1 * 4.43 = (1 + 2) v₂
v₂ = 4.43/3
v₂ = 1.48 m/s

(c) 2.98
E_i / E_f
(1/2 m₁ v₁²) / (1/2 (m₁ + m₂) v₂²)
= 9.81 / 3.29 = 2.98

(d) θ = 19.3°
1/2 (m₁ + m₂) v² + 0 = 0 + (m₁ + m₂) gh
h = v²/(2g) = 1.48²/19.6 = 0.112m
h = L - L cos(θ)
cos(θ) = (L - h)/2
θ = cos^-1((2-0.112)/2)
θ = 19.3°

(e) F_avg = -4440N
F_avg = I / t
= (P_f - P_i) / t
= (0 - (1+2)1.48) / 0.001
= -4440N

 

[BruceW]

I've gone through them, and they all look correct to me. Although part c should be equal to 3, your answer is slightly different due to rounding error. Part e is kind of vague, since the two masses end up being one mass, so its not like there is an impulse... But I think your answer is probably what they are looking for.