1. A cannon mounted on the back of a ship fires a 50.0 kg cannonball in the horizontal direction at a speed of 150 m/s. If the cannon and ship to which it is firmly attached have a mass of 4.00 x 103 kg and are initially at rest, what is the speed of the ship just after shooting the cannon. Ignore water resistance.
2. An 80.0-kg astronaut is floating at rest at a distance of 10.0 m from a spaceship when she runs out of oxygen and fuel to power her back to the spaceship. She removes her oxygen tank (3.0 kg) and flings it into space away from the ship with a speed of 15 m/s. At what speed does she recoil toward the spaceship?
3. A 2.10 x 103-lb sports car travelling east at 20.0 mi/h has a head-on collision with a 4.00 x 103-lb station wagon travelling west at 30.0 mi/h. If the vehicles remain locked together, what is their final velocity?
4. An 80.0=g arrow moving at 80.0 m/s hits and embeds in a 10.0-kg block resting on ice. Use the conservation of momentum principle to determine the speed of the block and arrow just after the collision.
1. In this problem, we can use the conservation of momentum principle. Initially, the momentum of the system (cannon, ship, and cannonball) is zero because everything is at rest. After the cannon is fired, the momentum of the cannonball and the recoiling ship should still add up to zero. So:
m_cannonball * v_cannonball = - m_ship * v_ship
where m_cannonball and m_ship are the masses of the cannonball and the ship (including the cannon), respectively, and v_cannonball and v_ship are their respective velocities after the cannon is fired. We want to find v_ship. Plugging in the given numbers:
(50.0 kg)(150 m/s) = - (4.00 x 10^3 kg) * v_ship
Solve for v_ship:
v_ship = - (50.0 kg)(150 m/s) / (4.00 x 10^3 kg) = -18.75 / 4 = -4.69 m/s
Since we have a negative sign in our answer, this means the ship is moving in the opposite direction of the fired cannonball. So the speed of the ship just after shooting the cannon is 4.69 m/s.
2. In this problem, we can again use the conservation of momentum principle. Initially, the momentum of the astronaut and her oxygen tank is zero because they are both at rest. After she throws the tank, the momentum of the astronaut and the thrown tank should still add up to zero. So:
m_tank * v_tank = - m_astronaut * v_astronaut
We want to find v_astronaut. Plugging in the given numbers:
(3.0 kg)(15 m/s) = - (80.0 kg) * v_astronaut
Solve for v_astronaut:
v_astronaut = - (3.0 kg)(15 m/s) / (80.0 kg) = -0.5625 m/s
Since we have a negative sign in our answer, this means the astronaut is moving in the opposite direction of the thrown tank, which is towards the spaceship. So the speed at which she recoils toward the spaceship is 0.5625 m/s.
3. To solve this problem we first need to convert the speeds from miles per hour to meters per second:
1 mi = 1609 m
1 hr = 3600 s
20.0 mi/h * (1609 m/mi) / (3600 s/hr) = 8.94 m/s for the sports car
30.0 mi/h * (1609 m/mi) / (3600 s/hr) = 13.4 m/s for the station wagon
Next, convert the weights from pounds to kilograms:
1 lb = 0.4536 kg
2.10 x 10^3 lb * 0.4536 kg/lb = 953 kg for the sports car
4.00 x 10^3 lb * 0.4536 kg/lb = 1814 kg for the station wagon
Now we can use the conservation of momentum principle:
p_initial = p_final
m1 * v1 + m2 * v2 = (m1 + m2) * v_final
where m1 and m2 are the masses of the sports car, and station wagon, respectively, and v1 and v2, are their respective initial velocities. Plugging in the numbers:
(953 kg)(8.94 m/s) + (1814 kg)(-13.4 m/s) = (953 kg + 1814 kg) * v_final
Solve for v_final:
v_final = [(953 kg)(8.94 m/s) + (1814 kg)(-13.4 m/s)] / (953 kg + 1814 kg) = -6.26 m/s
Since we have a negative sign in our answer, this means the final velocity of the vehicles is in the direction of the station wagon's initial velocity, which is westward. So their final velocity is 6.26 m/s westward.
4. To solve this problem, we can use the conservation of momentum principle:
p_initial = p_final
m_arrow * v_arrow + m_block * v_block = (m_arrow + m_block) * v_final
where m_arrow and m_block are the masses of the arrow and the block, respectively, and v_arrow and v_block are their respective initial velocities (v_block is initially at rest, so its velocity is 0). We want to find v_final. First, convert the mass of the arrow from grams to kilograms:
80.0 g = 80.0 / 1000 = 0.080 kg
Now, plug in the numbers:
(0.080 kg)(80.0 m/s) + (10.0 kg)(0 m/s) = (0.080 kg + 10.0 kg) * v_final
Solve for v_final:
v_final = (0.080 kg)(80.0 m/s) / (0.080 kg + 10.0 kg) = 0.639 m/s
So the speed of the block and arrow just after the collision is 0.639 m/s.
To solve these problems, we need to use the principle of conservation of momentum. According to this principle, the total momentum of an isolated system remains constant before and after a collision.
1. In this scenario, the cannon and ship can be treated as an isolated system. Initially, the ship is at rest, so its momentum is zero. The momentum of the cannonball is given by the product of its mass and velocity:
Momentum of cannonball = mass of cannonball * velocity of cannonball
= 50.0 kg * 150 m/s
= 7500 kg·m/s
Now, after the cannonball is fired, the cannon and ship will experience an equal and opposite recoil. Let the velocity of the ship after firing be V.
According to the conservation of momentum, the change in momentum of the cannonball (7500 kg·m/s) should equal the change in momentum of the ship and cannon system (-V * 4.00 x 10^3 kg):
Change in momentum of cannonball = change in momentum of ship and cannon system
7500 kg·m/s = -V * 4.00 x 10^3 kg
Solving for V:
V = (7500 kg·m/s) / (4.00 x 10^3 kg)
V ≈ -1.875 m/s
Note: The negative sign indicates that the ship moves in the opposite direction to the cannonball.
Therefore, the speed of the ship just after shooting the cannon is approximately 1.875 m/s in the opposite direction of the cannonball.
2. In this situation, the astronaut and the oxygen tank can be considered an isolated system. Initially, the astronaut is at rest, so their momentum is zero. The momentum of the oxygen tank is given by:
Momentum of oxygen tank = mass of oxygen tank * velocity of oxygen tank
= 3.0 kg * 15 m/s
= 45 kg·m/s
After the astronaut flings the tank, they will experience a recoil. Let the velocity of the astronaut after flinging the tank be V.
According to the conservation of momentum:
Change in momentum of oxygen tank = change in momentum of astronaut
45 kg·m/s = -V * 80.0 kg
Solving for V:
V = (45 kg·m/s) / (80.0 kg)
V ≈ -0.5625 m/s
The negative sign indicates that the astronaut moves in the opposite direction to the oxygen tank.
Therefore, the astronaut recoils with a speed of approximately 0.5625 m/s in the opposite direction.
3. In this case, the two vehicles colliding head-on can be treated as an isolated system. The initial momenta of the sports car and the station wagon are given by:
Momentum of sports car = mass of sports car * velocity of sports car
= 2.10 x 10^3 lb * (20.0 mi/h * 5280 ft/mi * 0.3048 m/ft) / 3600 s/h
Momentum of station wagon = mass of station wagon * velocity of station wagon
= 4.00 x 10^3 lb * (30.0 mi/h * 5280 ft/mi * 0.3048 m/ft) / 3600 s/h
Now, when the vehicles collide, they remain locked together. Let the final velocity of the system be V.
According to the conservation of momentum:
Initial total momentum = Final total momentum
Momentum of sports car + Momentum of station wagon = (mass of sports car + mass of station wagon) * V
Solving for V:
V = (Momentum of sports car + Momentum of station wagon) / (mass of sports car + mass of station wagon)
Note: Convert the masses and velocities to SI units before calculation.
Therefore, we can determine the final velocity by substituting the values.
4. In this question, we use the conservation of momentum principle once again. Initially, the arrow and the block are separate isolated systems. The momentum of the arrow is given by:
Momentum of arrow = mass of arrow * velocity of arrow
= 80.0 g * 80.0 m/s
Now, after the arrow embeds into the block, they become a combined system. Let the final velocity of the block and arrow system be V.
According to the conservation of momentum:
Initial momentum of arrow = Final momentum of block and arrow
Mass of arrow * velocity of arrow = (mass of arrow + mass of block) * V
Solving for V:
V = (Mass of arrow * velocity of arrow) / (mass of arrow + mass of block)
Note: Convert the mass of the arrow to kg before calculation.
By substituting the given values, we can find the speed of the block and arrow system just after the collision.