A 15.0 kg block is attached to a very light horizontal spring of force constant 575 N/m and is resting on a smooth horizontal table. Suddenly it is struck by a 3.00 kg stone traveling horizontally at 8.00 m/s to the right. After the collision clay sticks to the blick. Find max distance that spring will be compressed after the collision
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A 15.0 kg block is attached to a very light horizontal spring of force constant 575 N/m and is resting on a smooth horizontal table. Suddenly it is struck by a 3.00 kg stone traveling horizontally at 8.00 m/s to the right. After the collision clay sticks to the blick. Find max distance that spring will be compressed after the collision
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- A 2.00-kg pendulum ball is attached to a light-uniform cable of length 1.20 m that is hanging from the ceiling at an angle of 50.00 to the vertical. When the pendulum ball is released from the rest position, it swings to the very bottom of the motion where it collides head-on with a 1.40-kg block that is initially at rest on a level surface. If the pendulum ball recoils at 1.50 m/s, what is the speed of the block after the collision? a. 3.88 m/s b. 6.29 m/s c. 3.03 m/s d. 4.75 m/s e. 5.98 m/sWe can use conservation of momentum principles to compare the velocity of the marble before the collision to the velocity of the pendulum bob (with embedded marble) after the collision. After the collision, we can use conservation of energy principles to compare the initial kinetic energy of the (marble + pendulum) system with its potential energy when the pendulum reaches a maximum swing angle. Analyze the problem "backwards." Start with the motion of the (marble + pendulum) system after the inelastic collision. Using conservation of mechanical energy, derive an expression for the initial velocity of the marble + pendulum system as it starts to move. . Using trigonometry, you should be able to show that , where is the length of the ballistic pendulum, and is the maximum angle reached by the pendulum after the impact (show your work). Thus, the speed after the collision, , isA 26-g rifle bullet traveling 240 m/s embeds itself in a 3.7-kg pendulum hanging on a 2.6-m-long string, which makes the pendulum swing upward in an arc. (Figure 1) a)Determine the vertical component of the pendulum's maximum displacement. b)Determine the horizontal component of the pendulum's maximum displacement.
- A 1.9-kg ball is attached to the end of a 0.6-m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point of its swing, when it is moving horizontally, the ball collides with a 0.5-kg block initially at rest on a horizontal frictionless surface. The speed of the block just after the collision is 2.5 m/s. What is the speed of the ball just after the collision?A 5.00-g bullet moving with an initial speed of vo = 360 m/s is fired into and passes through a 1.00-kg block, as in the figure below. The block, initially at rest on a frictionless horizontal surface, is connected to a spring with a spring constant of 885 N/m. Vo www 5.00 cm- (a) If the block moves 5.00 cm to the right after impact, find the speed at which the bullet emerges from the block. m/s (b) If the block moves 5.00 cm to the right after impact, find the mechanical energy lost in the collision.A 2.0-kg object moves to the right with a speed of 4.0 m/s. It collides in a perfectly elastic head-on collision with a 5.0-kg object at rest. (a) What is the total kinetic energy after the collision? (b) What is the speed of the 2.0-kg object after the collision? (c) What is the speed of the 5.0-kg object after the collision?
- A bullet of mass m=5g moving with an initial speed 400m/s of 400 m/s is fired into and passes through a M=1 kg block, as in Figure. The block, initially at rest on frictionless, horizontal surface is connected to a 5.00cm! spring of force constant k=900 N/m. If the block moves d= 5 cm to the right after impact and the object stops. What is the speed of block at which the bullet emerges from it? 1,5 m/s 200 m/s 2 m/s 100 m/s 150 m/sA 5.00-g bullet moving with an initial speed of v; = 460 m/s is fired into and passes through a 1.00-kg block as shown in the fiqure below. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 910 N/m. The block moves d = 4.40 cm to the right after impact before being brought to rest by the spring. (a) Find the speed at which the bullet emerges from the block. m/s (b) Find the amount of initial kinetic energy of the bullet that is converted into internal energy in bullet-block system during the collision.A 0.430 kg block of wood rests on a horizontal frictionless surface and is attached to a spring (also horizontal) with a 26.0 N/m force constant that is at its equilibrium length. A 0.0600 kg wad of Play-Doh is thrown horizontally at the block with a speed of 2.60 m/s and sticks to it. Determine the amount in centimeters by which the Play-Doh-block system compresses the spring.
- Example 13-12 depicts the following secnario. A 0.990kg block slides on a frictionless, horizontal surface with a speed of 1.38 m/s. The block encounters an unstretched spring with a force constant of 218 N/m, as shown in the sketch. Before the block comes to rest, the spring is compressed by 9.30 cm .Consider two cars that can roll on a frictionless track. Initially, both cars are at rest. One car is launched from rest with a spring, and the cars undergo a completely inelastic collision. The spring constant is 850 N/m, and the spring is initially compressed by 0.075 m. The first car has a mass of 0.50 kg and the second has a mass of 1.0 kg. An identical spring is located to the right of the cars. a. The velocity of the first car just before collision. b. The velocities of the cars after collision. c. The distance the spring on the right will be compressed when the cars are completely stopped by the spring.10 kg - mass 500 N/m - k 0.2 m = x 30° = θ (theta) 0.2 = μk REQUIRED d = ?