1. A stone throw from ground level returns to the same level 4 s after. With what speed was the stone throw? Take g = 10m per square.
2. A cart is moving horizontally along a straight line with constant speed ofb30 m/s. A projectile is fired from the moving cart in such a way that it will return to the cart after the cart has moved 80 m. At what speed (relative to the cart) and at what angle (to the horizontal) must the projectile be fired?
3. A 10-g bullet of unknown speed is shot horizontally into a 2-kg block of wood suspended from the celling by a cord. The bullet hits the block and becomes lodged in it. After the collision, the block and the bullet swing to a height 30cm above the original position. What was the speed of the bullet? (This device is called the ballistic pendulum). Take g= 9.8m/s square.
1. V = Vo + g*Tr = 0
The rise time(Tr) = 1/2 of time in air.
Vo = -g*Tr = 10(4/2)= 20 m/s
1. To find the speed at which the stone was thrown, we can use the equation of motion: d = ut + (1/2)gt^2, where d is the distance traveled, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.
In this case, the stone returns to the same level after 4 seconds, and we assume the starting point is ground level. Since the stone returns to the same level, the distance traveled is 0. Thus, the equation becomes 0 = u(4) + (1/2)(10)(4^2).
Simplifying the equation, we get 0 = 4u + (1/2)(10)(16), which further simplifies to 0 = 4u + 80. Rearranging the equation, we find that 4u = -80, and dividing both sides by 4, we get u = -20 m/s.
Therefore, the speed at which the stone was thrown is 20 m/s (since speed is defined as the magnitude of velocity).
2. To find the speed and angle at which the projectile must be fired relative to the cart, we can use the principle of conservation of momentum.
Since there is no external horizontal force acting on the system, the horizontal momentum will be conserved. Initially, only the cart is moving horizontally with a constant speed of 30 m/s. When the projectile is fired, it also moves horizontally while rising and falling vertically. The projectile's horizontal momentum must be equal to the cart's initial momentum, which is the mass of the cart multiplied by its velocity.
Let's assume the mass of the projectile is m, its speed relative to the cart is v, and the angle it is fired at (relative to the horizontal) is θ.
The horizontal component of the projectile's velocity is v * cos θ.
Using conservation of momentum, we have:
(mass of cart * 30) = m * (v * cos θ)
Now, to find the speed at which the projectile must be fired relative to the cart, we need another equation. Since the projectile must return to the cart after the cart moves 80 m horizontally, the total horizontal distance traveled by the projectile will be 80 m.
The horizontal motion of the projectile can be described by the equation: distance = velocity * time
Considering that the initial velocity of the projectile is v (relative to the cart) and the time taken to travel 80 m horizontally is 80 / (v * cos θ), the equation becomes:
80 = v * cos θ * (80 / (v * cos θ))
Simplifying, we get:
80 = 80
This equation does not provide any information about the speed or angle of the projectile. Therefore, we cannot determine the specific values for v and θ using this approach.
3. To find the speed of the bullet, we can use the principle of conservation of momentum and the principle of conservation of mechanical energy.
Before the collision, the bullet has a mass of 10 g, and its speed is unknown. The block has a mass of 2 kg and is initially at rest. After the collision, the bullet becomes lodged in the block, and the block and bullet together swing to a height 30 cm above the original position.
Let's assume the final combined mass of the block and bullet is M. The bullet's speed before the collision is v, and the height reached after the collision is h.
According to the principle of conservation of momentum, the momentum before the collision is equal to the momentum after the collision.
The momentum before the collision is given by:
momentum_before = mass_bullet * velocity_bullet
The momentum after the collision is given by:
momentum_after = (mass_block + mass_bullet) * velocity_combined
Therefore, we have:
mass_bullet * velocity_bullet = (mass_block + mass_bullet) * velocity_combined
To find the speed of the bullet, we need to express velocity_combined in terms of variables we know. Using the principle of conservation of mechanical energy, we can relate the heights of both the block and bullet before and after the collision.
The gravitational potential energy of the system before the collision is zero, as both the bullet and block are at ground level.
The gravitational potential energy of the system after the collision is given by:
potential_energy_after = (mass_block + mass_bullet) * g * h
Applying the conservation of mechanical energy, we have:
potential_energy_after = (1/2) * (mass_block + mass_bullet) * velocity_combined^2
Simplifying the equation, we get:
g * h = (1/2) * velocity_combined^2
Rearranging the equation, we find that:
velocity_combined = (2 * g * h)^0.5
Substituting this value of velocity_combined back into the conservation of momentum equation, we have:
mass_bullet * velocity_bullet = (mass_block + mass_bullet) * (2 * g * h)^0.5
Finally, rearranging the equation and solving for the bullet's velocity, we find that:
velocity_bullet = [(mass_block + mass_bullet) * (2 * g * h)^0.5] / mass_bullet
Now, substituting the given values of mass_bullet, mass_block, g, and h, we can calculate the speed of the bullet.