Physics I: Tough one-dimenional kinematics problem.

This is about about as tough as one-dimensional kinematics problems come.

The problem: A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0s later. What was the rocket's acceleration?

(the picture is not given, it is my drawing)

Notice that the bolt does not just fall start free falling, it has the same upward velocity as the rocket, so is more like a projectile straight upward.

The important part of this one is to pay very careful attention to what your variables are when solving this algebraically. Make sure to note what variables you know, what you don't know, and what you need to find out. In this case, we are solving for a_R.

Known:

t₀ = 0.0s

t₁ = 4.0s

t₂ = 10.0s

y_0R = 0.0m

y_0B = y_1R

v_0B = v_1R

I start by writing up some kinematic equations for both the rocket and the bolt.

y_1B – y_0B = v_0B (t₂ – t₁) – ½ g (t₂ – t₁)²

In this case, (y_1B – y_0B) is not zero, and Δt is not equal to t₂! Now what? We don’t know v_0B or y_1B, so this problem is unsolvable, right? Wrong!

Now consider the equations for the rocket:

v_1R = v_0R + a_R t₁

v_1R = a_R t₁

y_1R = y_0R + v_0R t₁ + ½ a_R t₁

y_1R = ½ a_R t₁²

Notice that not only have we solved for both of the variables in our bolt’s equation, but they are both in terms of acceleration, meaning we’ll only have 1 variable now. Back to the bolt’s equation:

y_1B – y_0B = v_0B (t₂ – t₁) – ½ g (t₂ – t₁)²

– y_0B = v_0B t₁ – ½ g (t₂ – t₁)² (Notice that v_0B = v_1R = a_R t₁, and y_0B = y_1R = ½ a_R t₁²)

– ½ a_R t₁² = a_R t₁ (t₂ – t₁) – ½ g (t₂ – t₁)²

Solve for a_R.

a_R t₁ (t₂ – t₁)  + ½ a_R t₁² = ½ g (t₂ – t₁)²

a_R (t₁ (t₂ – t₁)  + ½ t₁²) = ½ g (t₂ – t₁)²

a_R = ½ g (t₂ – t₁)²/ (t₁ (t₂ – t₁)  + ½ t₁²)

Solution: a_R = 5.5m/s²

Physics I: Projectile motion on a ramp.

Here's a good one.

In one contest at the country fair, a spring-loaded plunger launches a ball at a speed of 3.0 m/s from one corner of a smooth, flat board that is tilted up at a 20 degree angle. To win, you must make the ball hit a small target at the adjacent corner, 2.50 m away. At what angle should you tilt the ball launcher?

The ball is rolling up and down the ramp, from one corner to another. This really isn't much different from a standard projectile motion problem, the only difference is the y component acceleration is -gsin20 rather than simply -g.

A few things to note: we have the initial velocity but not its components, so this is quite do-able. Be careful not to mix the slope of the ramp and the unknown angle we're trying to find!

v_0x = v₀cosθ
v_0y = v₀sinθ 

Solve for t when v_y = 0 (the peak of the trajectory).

v_y = v_0y + a_yt = v₀sinθ - gtsin20
gtsin20 = v₀sinθ (plugging in v₀sinθ for v_0y)
t = (v₀sinθ)/gsin20

Remember, this is the time when the ball is at its highest point, make sure not to forget that we're using x₁ (2.50 m / 2), not x₂ (the final position).

x = v₀t
x₁ = v_0xt₁
x₁ = 2v_0xt₁
x₁ = v_0x(v₀sinθ/gsin20)
x₁ = v₀²cosθsinθ/gsin20 (plugging in v₀cosθ for v_0x)
x₁ = v₀²sin2θ/2gsin20 (using double angle formula 2sinθcosθ = sin2θ, since we didn't have a 2 I multiplied the whole fraction by 2/2)
sin2θ = 2_x1gsin20/v₀²
sin^-1(2_x1gsin20/ v₀²) = 2θ (reminder, plug in 1.25m for x₁, not 2.50, since this is x at t₁ which is at ½ t_f)
2θ = 68.6 degrees

Solution: θ = 34.3 degrees   
 

Physics I, Newton's Laws: Interesting pulley situation.

The problem: Find an expression for the acceleration of m₁ (the table is frictionless).

This is pretty straightforward, but it's easy to overlook something.

The acceleration in m₁ and m₂ are not equal. The block m₂ is accelerating half as fast as m₁. To understand this, draw a free-body diagram.

m₁ is pulled to the right by a tension force T. Since there is no friction force, the normal force and weight of m1 isn't important.

m₂ is pulled down by its weight, and pulled up by two tension forces, both equal to T.

a₁ = 2a₂
a₂ = a₁/2

F_{net (m1)} = T = m₁a₁
F_{net (m2)} = 2T - m₂g = m₂a₂

Since T = m₁a₁, plug that into the second expression. Also, since we're solving for a₁, eliminate a₂ by plugging in a₁/2

2m₁a₁ - m₂g = (m₂a₁)/2
4m₁a₁ - 2m₂g = m₂a₁
4m₁a₁ - m₂a₁ = 2m₂g
a₁(4m₁ - m₂) = 2m₂g

Solution: a₁ = (2m₂g)/(4m₁ - m₂)

Physics I: A painter on a chair attached to a pulley.

A house painter uses the chair and pulley arrangement of the figure to lift himself up the side of a house. The painter's mass is 70 kg and the chair's mass is 10 kg. With what force must he pull down on the rope in order to accelerate upward at 0.20 m/s²?

Your first instinct might be to guess that he simply needs to pull down as hard as his weight + the chair's weight plus enough force to accelerate. This is wrong! 

Start by drawing a free body diagram. The painter has a normal force N with the chair (upward), and a reaction force of the chair pushing up on him due to the tension pulling the chair up. He has a downward force equal to his weight. The chair has an upward tension force, a downward normal force from its contact with the painter (N), and its weight.

Fnet(painter) = N + T - W_p = m_pa

Fnet(chair) = T - N - W_c = m_ca

Since acceleration is known, and the painter's pull is equal to the tension force, we set acceleration equal to 0.20 m/s^2 and solve for T. This is a little bit more difficult since the normal force is not known (the normal force in this case is NOT equal to the painter's weight, because the painter is also in contact with the rope pulling downward and the chair is moving upwards!

Solve for both N and T algebraically first.

T = m_pa + W_p - N
N = T - W_c - m_ca

Now plug N into the T equation to eliminate the N variable.

T = m_pa + W_p - (T - W_c - m_ca)
T = m_pa + W_p -T + W_c + mca
2T = m_pa + m_pg  + m_cg + m_ca
T = (m_pa + m_ca + m_pg + m_cg) / 2

Solution: T = 400 N