1. A 10.0 g object moving to the right at 25.0 cm/s makes an elastic head-on collision with a 15.0 g object moving in the opposite direction at 36.0 cm/s. Find the velocity of each object after the collision.
10.0 g object:
15.0g object:
2. A dentist's drill starts from rest. After 3.00 s of constant angular acceleration it turns at a rate of 2.00 multiplied by 104 rev/min.
(a) Find the drill's angular acceleration.
(b) Determine the angle (in radians) through which the drill rotates during this period.
well since everything is in cm and seconds and grams we can stay in the centimeter, gram, second or"cgs" units.
Initial momentum = 10*25 - 15*36
= -290 (to the right is positive)
final momentum =same = -290 = 10 U + 15 V
elastic so initial KE = final KE
(1/2) 10 (625) + (1/2) 15 (1296)
3125+9720
=12,845 dynes
final KE =12,845 = 5 U^2 + 7.5 V^2
so two equations, two unknowns:
-290 = 10 U + 15 V
12,845 = 5 U^2 + 7.5 V^2
U = - (1.5 V + 29)
U^2 = 2.25 V^2 + 87 V + 841
12,845 = 5(2.25 V^2 + 87 V + 841) + 7.5 V^2
You can solve the quadratic I think
in 3 seconds from 0 to 2*10^4 revs/min
first change rev/min to radians/sec
2*10^4rev/min (1 min/60 s)(2 pi rad/rev)
=2094 radians/s
alpha = change in speed/change in time = 2094/3 = 698 radians/s^2
angle = (1/2) alpha t^2
=349*9 = 3141 radians
To solve the first question about the collision of the two objects:
1. First, we need to use the conservation of momentum to solve for the final velocities of the objects.
The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.
Mathematically, we can express this as:
m1 * v1_ini + m2 * v2_ini = m1 * v1_fin + m2 * v2_fin,
where m1 and m2 are the masses of the objects, v1_ini and v2_ini are their initial velocities, and v1_fin and v2_fin are their final velocities.
2. Given:
m1 = 10.0 g = 0.010 kg (converting grams to kilograms),
v1_ini = 25.0 cm/s = 0.25 m/s (converting cm/s to m/s),
m2 = 15.0 g = 0.015 kg,
v2_ini = -36.0 cm/s = -0.36 m/s.
3. Substituting the given values into the momentum conservation equation:
(0.010 kg * 0.25 m/s) + (0.015 kg * -0.36 m/s) = (0.010 kg * v1_fin) + (0.015 kg * v2_fin).
4. Solving for v1_fin and v2_fin, we find:
v1_fin = (-0.015 kg * -0.36 m/s + 0.015 kg * 0.25 m/s) / 0.010 kg,
v2_fin = (0.010 kg * 0.25 m/s - 0.010 kg * -0.36 m/s) / 0.015 kg.
Calculating the above equations gives us:
v1_fin = 0.48 m/s,
v2_fin = -0.56 m/s.
So, after the collision:
1. The 10.0 g object (m1) will have a final velocity of 0.48 m/s to the right.
2. The 15.0 g object (m2) will have a final velocity of -0.56 m/s to the left.
Now, let's move on to the second question about the dentist's drill:
(a) To find the angular acceleration of the drill, we can use the formula:
ω = ω0 + α * t,
where ω is the final angular velocity, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time.
Given:
ω0 = 0 rev/min,
ω = 2.00 * 10^4 rev/min = (2.00 * 10^4 rev/min) * (2π rad/1 rev) * (1 min/60 s) = 209.44 rad/s,
t = 3.00 s.
Rearranging the formula, we have:
α = (ω - ω0) / t = (209.44 rad/s - 0 rad/s) / 3.00 s.
Therefore, the drill's angular acceleration is 69.81 rad/s^2.
(b) To determine the angle (in radians) through which the drill rotates during this period, we can use the formula:
θ = ω0 * t + (1/2) * α * t^2,
where θ is the angle in radians.
Given:
ω0 = 0 rad/s,
t = 3.00 s,
α = 69.81 rad/s^2.
Substituting the values into the formula, we find:
θ = 0 rad/s * 3.00 s + (1/2) * 69.81 rad/s^2 * (3.00 s)^2.
Simplifying the equation gives us:
θ = 313.65 radians.
Therefore, the drill rotates through an angle of 313.65 radians during this period.