A ballistic pendulum is a simple, butpractical, device that can be used tomeasure the speed of a bullet. Thependulum consists of a block of woodwith mass 4.30 kg, at the end of a thinwire so that the center of the block is85.6 cm below the top of the wire.With the block at rest and the wirevertical, a bullet of mass 29.0 grams isfired at the block. It embeds in the blockand causes the block to swing so that itmomentarily comes to rest when thewire makes an angle of 35.80 with thevertical. Find the initial speed, in m/s, ofthe bullet.I

We are given the angle made by bullet pendulum system. We find the potential energy of the final…

Answered: A ballistic pendulum is a simple, but…

Similar questions

In walking the body goes through both a breaking impulse and a propulsion impulse. The breaking phase lasts .3 seconds and the propulsion phase lasts .33 seconds. The body goes through a decrease of the vertical velocity of .1 m/s as they come into the ground during the breaking impulse and an increase in the vertical velocity of .4 m/s during the propulsion. The individual has a weight of 700 newtons. What is the acceleration horizontally for both the breaking impulse and the propulsive impulse if the coefficient of friction is .32?
Two hockey players race for the puck in the middle of the rink. Player 1 has a mass of 69 kg and a velocity of 5.6 m/s at an angle of 32° with the rink board. Player 2 has a mass of 78 kg and a velocity of 7.1 m/s at an angle of 16° with the same board. The players become locked together in a completely inelastic collision. Find the energy in Joules lost in the collision.
One billiard ball (mass 118 grams) is moving due east at 5.0 m/s. A tennis ball (mass 321 grams) is moving due west at 4.0 m/s. The two balls have a glancing blow, which means they do not bounce along the same line as their original motion, but rather, deflect at an angle to their original paths. You measure the billard ball's motion as it moves 26.0° north of east, and you measure that the tennis ball is travelling 19.0° south of west. What is the speed of the billiard ball after the collision? Your Answer: Answer
The pendulum apparatus in the video has a total mass of 578 grams. The pendulum rod has a small mass of 20 grams so you can treat all the mass of the pendulum as if it were hanging at the end which is called the “bob”. The mass of the rod should be taken into account in determining the center of mass of the apparatus which is 71.3 cm (not the length of the pendulum). A marble which hass a mass of 10 g is fired at the pendulum bob which has clay on it which allows the marble to stick to it. Derive a formula for measuring the velocity of the marble (vm) as a function of the mass of the marble mm, the mass of the marble/bob mmb and the velocity of the marble and bob vmb using the conservation of momentum.
A large block of mass 3.80 kg slides on a frictionless table and collides head-on with a small block of mass 2.50 kg. After the collision both blocks move in the same direction as the original motion of the large block, and fall off the edge of the table. The small block hits the floor at a point 2.18 meters horizontally from the edge of the table, while the large block hits the floor at a point 1.38 meters horizontally from the edge of the table. The top of the table is 1.10 m above the floor. Find the speed, in m/s, of the large block before the collision.
In a ballistic pendulum experiment, a small marble is fired into a cup attached to the end of a pendulum. If the mass of the marble is 0.0235 kg and the mass of the pendulum is 0.250 kg, how high h will the pendulum swing if the marble has an initial speed of 5.65 m/s? Assume that the mass of the pendulum is concentrated at its end so that linear momentum is conserved during this collision. h = m
In the figure, projectile particle 1 is an alpha particle and target particle 2 is an oxygen nucleus. The alpha particle is scattered at angle 0₁ = 63.1° and the oxygen nucleus recoils with speed 1.60 × 105 m/s and at angle 0₂ = 52.0°. In atomic mass units, the mass of the alpha particle is 4.00 u and the mass of the oxygen nucleus is 16.0 u. What are the (a) final and (b) initial speeds of the alpha particle? (a) Number (b) Number i m₁ Vli Units Units mg vaf x
A steel ball of mass 0.600 kg is fastened to a cord that is 80.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal. At the bottom of its path, the ball strikes a 3.00 kg steel block initially at rest on a frictionless surface. The collision is elastic. (a) Find the speed of the ball just after collision. (No Response) m/s (b) Find the speed of the block just after collision. (No Response) m/s
A steel ball of mass 0.440 kg is fastened to a cord that is 80.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal, as shown in the figure. At the bottom of its path, the ball strikes a 3.10 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision. (a) Number i (b) Number Unit Unit
In a ballistic pendulum experiment, a small marble is fired into a cup attached to the end of a pendulum. If the mass of the marble is 0.0285 kg and the mass of the pendulum is 0.250 kg, how high h will the pendulum swing if the marble has an initial speed of 5.85 m/s? Assume that the mass of the pendulum is concentrated at its end so that linear momentum is conserved during this collision. h = m
You fire an old rifle at a block of wood. The mass of the block of wood is 231 g and it dimensions are 10.2 cm × 10.2 cm × 3.16 cm. The bullet has a mass of 24.6 grams. Assume no air resistance and that the collision is perfectly inelastic. The velocity of the bullet is 22: m/s right before it hits the block. Assume no mass is lost when the two objects collide. 10.2 cm 1. The distance the bullet hits from the center of mass is 3.4 cm COM. How high up does the block travel? cm 2. What is the rotation rate (rev/s) of the block & bullet system directly after impact? 3.16
Two 1.9 kg bodies, A and B, collide. The velocities before the collision are v A= (14î + 34 j) m/s and v = (-15î + 9.0j) m/s. After the collision, v = (-7.oî + 16 j) m/s. (a) What is the final velocity of B? m/s (b) What is the change in the total kinetic energy (including sign)? 121.69 J The net external force on the two particle system is zero; thus, the net momentum of the system cannot change. The net momentum along the x axis before the collision must equal that after the collision. Same for the y axis. Use the definition of kinetic energy to find the total K before the collision and the total K after.

SEE MORE QUESTIONS