Sample Problems

Solution:

a) First, let's find the maximum height reached above the top of the building:

\( h_{max} = \frac{v_0^2}{2g} = \frac{(15 \text{ m/s})^2}{2 \cdot 10 \text{ m/s}^2} = \frac{225}{20} = 11.25 \text{ m} \)

So the maximum height above the ground is:

\( H_{max} = 20 \text{ m} + 11.25 \text{ m} = 31.25 \text{ m} \)

b) To find the time to hit the ground, we can use the position equation:

\( y = y_0 + v_0t - \frac{1}{2}gt^2 \)

With \( y = 0 \) (ground), \( y_0 = 20 \text{ m} \) (initial height), \( v_0 = 15 \text{ m/s} \) (initial velocity), and \( g = 10 \text{ m/s}^2 \):

\( 0 = 20 + 15t - 5t^2 \)

\( 5t^2 - 15t - 20 = 0 \)

\( t^2 - 3t - 4 = 0 \)

Using the quadratic formula: \( t = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2} \)

\( t = 4 \text{ s} \) or \( t = -1 \text{ s} \)

Since we need the positive time, \( t = 4 \text{ s} \)

c) The velocity just before hitting the ground can be found using:

\( v = v_0 - gt = 15 - 10 \cdot 4 = 15 - 40 = -25 \text{ m/s} \)

The negative sign indicates the ball is moving downward, so the speed is 25 m/s.

Solution:

a) Let's define the positive direction as down the incline for the 2 kg block and downward for the 1 kg mass.

For the 2 kg block on the incline:

\( m_1g\sin\theta - T = m_1a \)

\( 2 \cdot 10 \cdot \sin30° - T = 2a \)

\( 10 - T = 2a \) ... (1)

For the 1 kg hanging mass:

\( m_2g - T = m_2a \)

\( 1 \cdot 10 - T = 1 \cdot a \)

\( 10 - T = a \) ... (2)

From equation (2): \( T = 10 - a \)

Substituting into equation (1):

\( 10 - (10 - a) = 2a \)

\( 10 - 10 + a = 2a \)

\( a = 2a \)

This is inconsistent, which means we made an error in our setup.

Let's reconsider: When the 1 kg mass moves down, the 2 kg block moves up the incline. So the accelerations are in opposite directions.

For the 2 kg block (positive direction up the incline):

\( T - m_1g\sin\theta = m_1a \)

\( T - 2 \cdot 10 \cdot \sin30° = 2a \)

\( T - 10 = 2a \) ... (1)

For the 1 kg hanging mass (positive direction downward):

\( m_2g - T = m_2a \)

\( 10 - T = a \) ... (2)

Since the string connects both masses, if one accelerates at rate a, the other also accelerates at rate a.

From equation (1): \( T = 2a + 10 \)

Substituting into equation (2):

\( 10 - (2a + 10) = a \)

\( -2a = a \)

\( a = 0 \)

This means the system is in equilibrium. Let's double-check our setup.

Actually, let's define the positive direction as down the incline for the block and downward for the hanging mass. The accelerations are in the same direction in this coordinate system.

For the 2 kg block:

\( m_1g\sin\theta - T = m_1a \)

\( 2 \cdot 10 \cdot 0.5 - T = 2a \)

\( 10 - T = 2a \) ... (1)

For the 1 kg hanging mass:

\( m_2g - T = m_2a \)

\( 10 - T = a \) ... (2)

From equations (1) and (2):

\( 10 - T = 2a \)

\( 10 - T = a \)

Therefore: \( 2a = a \), which is only true if \( a = 0 \)

Let's try once more with the correct approach. The key insight is that the accelerations of the two masses are related but not necessarily equal in magnitude.

If the hanging mass moves down by distance x, the block moves up the incline by the same distance x. The accelerations are related but in different directions.

For the 2 kg block (positive direction down the incline):

\( m_1g\sin\theta - T = m_1a_1 \)

\( 2 \cdot 10 \cdot 0.5 - T = 2a_1 \)

\( 10 - T = 2a_1 \) ... (1)

For the 1 kg hanging mass (positive direction downward):

\( m_2g - T = m_2a_2 \)

\( 10 - T = a_2 \) ... (2)

Since the string is inextensible, \( a_1 = a_2 = a \)

From equation (2): \( T = 10 - a \)

Substituting into equation (1):

\( 10 - (10 - a) = 2a \)

\( a = 2a \)

\( a = 0 \)

This still gives us a = 0, which means the system is in equilibrium. Let's check our calculation of the component of weight down the incline.

For a 30° incline: \( \sin30° = 0.5 \)

So the component of weight down the incline is: \( 2 \cdot 10 \cdot 0.5 = 10 \text{ N} \)

The weight of the hanging mass is: \( 1 \cdot 10 = 10 \text{ N} \)

Since these forces are equal, the system is indeed in equilibrium with \( a = 0 \) and \( T = 10 \text{ N} \).

b) The tension in the string is \( T = 10 \text{ N} \)

Solution:

a) We can use conservation of energy to find the speed at the lowest point.

At the initial position (60° from vertical):

Initial height: \( h_i = L - L\cos\theta = L(1-\cos\theta) = 1.0 \cdot (1-\cos60°) = 1.0 \cdot (1-0.5) = 0.5 \text{ m} \)

Initial energy: \( E_i = mgh_i = 0.5 \cdot 10 \cdot 0.5 = 2.5 \text{ J} \)

At the lowest point:

Height: \( h_f = 0 \)

Final energy: \( E_f = \frac{1}{2}mv^2 \)

By conservation of energy: \( E_i = E_f \)

\( mgh_i = \frac{1}{2}mv^2 \)

\( v^2 = 2gh_i = 2 \cdot 10 \cdot 0.5 = 10 \)

\( v = \sqrt{10} \approx 3.16 \text{ m/s} \)

b) At the lowest point, the tension in the rod has two components:

1. The component needed to provide the centripetal force: \( F_c = \frac{mv^2}{r} = \frac{0.5 \cdot 10}{1.0} = 5 \text{ N} \)

2. The component to balance the weight: \( F_g = mg = 0.5 \cdot 10 = 5 \text{ N} \)

Total tension: \( T = F_c + F_g = 5 + 5 = 10 \text{ N} \)

Solution:

Let's define our coordinate system with the x-axis along the initial direction of the 2 kg object.

Initial momentum:

\( \vec{p}_i = m_1\vec{v}_{1i} + m_2\vec{v}_{2i} = 2 \cdot 3\hat{x} + 0 = 6\hat{x} \text{ kg·m/s} \)

After the collision, let's denote:

\( \vec{v}_{1f} = v_{1f}\cos\theta_1\hat{x} + v_{1f}\sin\theta_1\hat{y} \) (for the 2 kg object)

\( \vec{v}_{2f} = v_{2f}\cos\theta_2\hat{x} + v_{2f}\sin\theta_2\hat{y} \) (for the 1 kg object)

We know that \( \theta_1 = 30° \)

By conservation of momentum:

x-component: \( 6 = 2v_{1f}\cos30° + v_{2f}\cos\theta_2 \)

y-component: \( 0 = 2v_{1f}\sin30° + v_{2f}\sin\theta_2 \)

From the y-component: \( v_{2f}\sin\theta_2 = -2v_{1f}\sin30° = -2v_{1f} \cdot 0.5 = -v_{1f} \)

This means \( \sin\theta_2 < 0 \), so \( \theta_2 \) is in the fourth quadrant (i.e., \( \theta_2 < 0 \)).

For an elastic collision, kinetic energy is also conserved:

\( \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 \)

\( \frac{1}{2} \cdot 2 \cdot 3^2 + 0 = \frac{1}{2} \cdot 2 \cdot v_{1f}^2 + \frac{1}{2} \cdot 1 \cdot v_{2f}^2 \)

\( 9 = v_{1f}^2 + \frac{1}{2}v_{2f}^2 \)

We now have three equations:

1. \( 6 = 2v_{1f}\cos30° + v_{2f}\cos\theta_2 \)

2. \( 0 = 2v_{1f}\sin30° + v_{2f}\sin\theta_2 \)

3. \( 9 = v_{1f}^2 + \frac{1}{2}v_{2f}^2 \)

From equation 2: \( v_{2f}\sin\theta_2 = -v_{1f} \)

For elastic collisions between two objects, we can use the fact that the relative velocity along the line of centers is reversed:

\( v_{1f}\cos\theta_1 - v_{2f}\cos\theta_2 = -(v_{1i} - v_{2i}) = -3 \)

Substituting \( \cos30° = \frac{\sqrt{3}}{2} \):

\( v_{1f} \cdot \frac{\sqrt{3}}{2} - v_{2f}\cos\theta_2 = -3 \)

\( v_{1f} \cdot \frac{\sqrt{3}}{2} + 3 = v_{2f}\cos\theta_2 \)

From momentum conservation (x-component):

\( 6 = 2v_{1f} \cdot \frac{\sqrt{3}}{2} + v_{2f}\cos\theta_2 \)

\( 6 = v_{1f}\sqrt{3} + v_{2f}\cos\theta_2 \)

Combining these equations:

\( 6 = v_{1f}\sqrt{3} + v_{1f} \cdot \frac{\sqrt{3}}{2} + 3 \)

\( 3 = v_{1f}\sqrt{3} + v_{1f} \cdot \frac{\sqrt{3}}{2} \)

\( 3 = v_{1f}\sqrt{3}(1 + \frac{1}{2}) \)

\( 3 = v_{1f}\sqrt{3} \cdot \frac{3}{2} \)

\( v_{1f} = \frac{3}{\sqrt{3} \cdot \frac{3}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \approx 1.155 \text{ m/s} \)

a) The speed of the 2 kg object after the collision is approximately 1.155 m/s.

Now we can find \( v_{2f} \) using the conservation of momentum:

\( 6 = 2 \cdot 1.155 \cdot \frac{\sqrt{3}}{2} + v_{2f}\cos\theta_2 \)

\( 6 = 1.155\sqrt{3} + v_{2f}\cos\theta_2 \)

\( 6 - 1.155\sqrt{3} = v_{2f}\cos\theta_2 \)

From equation 2: \( v_{2f}\sin\theta_2 = -1.155 \)

Using these two equations:

\( (v_{2f}\cos\theta_2)^2 + (v_{2f}\sin\theta_2)^2 = v_{2f}^2 \)

\( (6 - 1.155\sqrt{3})^2 + (-1.155)^2 = v_{2f}^2 \)

Calculating: \( v_{2f} \approx 4.04 \text{ m/s} \)

For the angle:

\( \sin\theta_2 = \frac{-1.155}{4.04} \approx -0.286 \)

\( \theta_2 \approx -16.6° \)

b) The 1 kg object moves at approximately 4.04 m/s at an angle of -16.6° (or 343.4° if measured counterclockwise from the positive x-axis).

Solution:

a) To find the angular acceleration, we need to use the rotational form of Newton's second law:

\( \tau = I\alpha \)

The torque due to gravity acts at the center of mass, which is at a distance L/2 from the pivot:

\( \tau = Mg \cdot \frac{L}{2} \cdot \cos\theta \)

At the moment of release, \( \theta = 0° \), so \( \cos\theta = 1 \)

\( \tau = Mg \cdot \frac{L}{2} \)

The moment of inertia of a uniform rod about one end is:

\( I = \frac{1}{3}ML^2 \)

Now we can find the angular acceleration:

\( \alpha = \frac{\tau}{I} = \frac{Mg \cdot \frac{L}{2}}{\frac{1}{3}ML^2} = \frac{3g}{2L} \)

b) The linear acceleration of the center of mass is:

\( a_{cm} = \alpha \cdot \frac{L}{2} = \frac{3g}{2L} \cdot \frac{L}{2} = \frac{3g}{4} \)

c) The linear acceleration of the free end is:

\( a_{end} = \alpha \cdot L = \frac{3g}{2L} \cdot L = \frac{3g}{2} \)

Solution:

First, let's use conservation of energy to relate the initial and final states.

Initial energy (at the rim):

Potential energy: \( U_i = mgR \) (taking the bottom of the bowl as the zero level)

Kinetic energy: \( K_i = 0 \) (released from rest)

Total initial energy: \( E_i = mgR \)

Final energy (at the bottom):

Potential energy: \( U_f = 0 \)

Kinetic energy: \( K_f = \frac{1}{2}mv^2 \)

Total final energy: \( E_f = \frac{1}{2}mv^2 \)

By conservation of energy: \( E_i = E_f \)

\( mgR = \frac{1}{2}mv^2 \)

\( v^2 = 2gR \)

Now, at the bottom of the bowl, the block is moving in a circular path with radius R. The normal force has two components:

1. A component to provide the centripetal force: \( F_c = \frac{mv^2}{R} \)

2. A component to balance the weight: \( F_g = mg \)

The total normal force is the vector sum of these components. At the bottom of the bowl, both forces act in the same direction (upward), so:

\( N = F_c + F_g = \frac{mv^2}{R} + mg \)

Substituting \( v^2 = 2gR \):

\( N = \frac{m \cdot 2gR}{R} + mg = 2mg + mg = 3mg \)

Therefore, the normal force at the bottom of the bowl is \( N = 3mg \).