1. A 3.25 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60.0° with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.195 s, what is the average force exerted on the ball by the wall?

x component: N
y component: N

2. A 7.0 g bullet is fired into a 1.5 kg ballistic pendulum. The bullet emerges from the block with a speed of 200 m/s, and the block rises to a maximum height of 11 cm. Find the initial speed of the bullet.

1. The velocity component normal to the wall (which I will call the x direction) changes from 10.0 sin 60 to -10.0 sin 60. That is a velocity component change of 17.32 m/s. Multiply that by the mass for the momentum change in that direction. Dividing by the contact time gives you the x xomponent of the force.

The velocity component in the y direction does not change, so there is no applied force in that direction.

2. Get the velocity V of the block right after it is hit by applying conservation of energy to the rise of the ballistic pendulum afterwards.
V = sqrt (2gH),
here H = 0.11 m and g = 9.8 m/s^2.

Let v1 and v2 be the velocity of the bullet before and after passing through the block. Apply conservation of linear momentum to that impact
m (v1 - v2) = M V
M is the mass of the block amd m is the mass of the bullet.

Solve for v1

A spy in a speed boat is being chased down a river by government officials in a faster craft. Just as the officials' boat pulls up next to the spy's boat, both boats reach the edge of a 4.8 m waterfall. The spy's speed is 16 m/s and the officials; speed is 26 m/s. How far apart will the two vessels be when they land below the waterfall? The acceleration of gravity is 9.81 m/s . Answer in units of m

1. To find the average force exerted on the ball by the wall, we need to use the impulse-momentum principle. Impulse is defined as the change in momentum of an object, which can be calculated by multiplying the average force exerted on the object by the time it is in contact with the wall.

First, let's calculate the initial momentum of the ball before it strikes the wall. Momentum is defined as the product of mass and velocity. The mass of the ball is 3.25 kg, and the velocity is 10.0 m/s at an angle of 60.0° with the surface. To calculate the x and y components of velocity, we can use trigonometry.

The x-component velocity (Vx) can be calculated by multiplying the total velocity (10 m/s) by the cosine of the angle (60°).

Vx = 10 m/s * cos(60°)
Vx = 10 m/s * 0.5
Vx = 5 m/s

The y-component velocity (Vy) can be calculated by multiplying the total velocity (10 m/s) by the sine of the angle (60°).

Vy = 10 m/s * sin(60°)
Vy = 10 m/s * 0.866
Vy = 8.66 m/s

Now, we can find the initial momentum (p) by multiplying the mass (m) by the total velocity (v).

p = m * v
p = 3.25 kg * 10 m/s
p = 32.5 kg·m/s

When the ball bounces off, it has the same speed and angle, but the direction of the velocity vector changes. Therefore, the change in momentum is equal to the initial momentum (since the mass remains the same).

Now, let's calculate the average force exerted on the ball by the wall. We can use the formula:

Impulse = change in momentum
Average force * time = Delta p

Average force = Delta p / time

Since the change in momentum is equal to the initial momentum, and the time is given as 0.195 s:

Average force = 32.5 kg·m/s / 0.195 s
Average force ≈ 166.67 N (rounded to two decimal places)

Therefore, the average force exerted on the ball by the wall is approximately 166.67 N.

2. To find the initial speed of the bullet, we can use the principle of conservation of momentum. The law of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this case, since the bullet collides with the ballistic pendulum and no external forces are mentioned, we can assume that momentum is conserved.

The initial momentum of the system is the sum of the bullet's momentum and the pendulum's momentum. We can calculate the initial momentum (p_initial) by multiplying the mass (m) by the initial velocity (v_initial).

p_initial = m_bullet * v_bullet + m_pendulum * v_pendulum

Given:
m_bullet = 7.0 g = 0.007 kg (convert grams to kilograms)
v_bullet = 200 m/s
m_pendulum = 1.5 kg
v_pendulum = 0 m/s (at rest initially)

p_initial = (0.007 kg * 200 m/s) + (1.5 kg * 0 m/s)
p_initial = 1.4 kg·m/s

Since momentum is conserved, the initial momentum (p_initial) is equal to the final momentum (p_final), which is just the momentum of the bullet after it has emerged from the block.

p_initial = p_final
1.4 kg·m/s = m_bullet * v_final

We need to solve for v_final to find the initial speed of the bullet. We know that the bullet's mass (m_bullet) and the final speed (v_final) are given.

Given:
m_bullet = 0.007 kg
v_final = 200 m/s

1.4 kg·m/s = 0.007 kg * v_final
v_final = 1.4 kg·m/s / 0.007 kg
v_final ≈ 200 m/s

Therefore, the initial speed of the bullet is approximately 200 m/s.