a) The center mass position is:
b) We use the conservation of linear mometum:
c) we calculate the initial kinetic energy:
final kinetic energy:
d) we have to apply the conservation of linear mometum again:
using given data we get:
also we know that this is a elastic collision:
so:
solving the system we have:
e) we have:
solving for Vcannon:
Problem 2 A 1.0 kg cannonball is fired horizontally with a speed of 200 rn/s at...
refer to picture. refer to pic Two blocks of masses 0.990 kg and 2.00 kg are at rest on a frictionless surface. A 10.0 g bullet is fired horizontally at a speed of 750 m/s and embeds itself inside the smaller block, as shown. The block with the bullet then collides elastically with the larger block. Determine, the speed of the smaller block after the bullet embeds itself inside. a. the amount of thermal energy produced. b. the final speed...
A bullet with a mass of 1.60 g is fired horizontally at two blocks resting on a smooth and frictionless table top as shown in the Figure. The bullet passes trhough the first 1.60 kg block, and embeds itself in a second 1.80 kg block. Speeds v1 = 1.60 m/s and v2 = 4.60 m/s, are thereby imparted on the blocks. The mass removed from the first block by the bullet can be neglected. Find the speed of the bullet...
Block A of a mass 0.7 kg is sliding to the right at a speed of 4.7 m/s while block B of mass 4.5kg is sliding to the right with a velocity of 1.2 m/s. The surface is frictionless for both blocks. If they collide perfectly elastically what is the speed of block A after the collision?
A bullet of mass 0.056 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 1.5 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? v= m/s (b) Calculate the kinetic energy of the bullet plus the block before the collision: K; = (c) Calculate the kinetic energy of the bullet plus the block after the...
A bullet of mass 0.017 kg traveling horizontally at a high speed of 210 m/s embeds itself in a block of mass 4 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? Vf = m/s ) Calculate the total translational kinetic energy before and after the collision. Ktrans,i = Ktrans,f = (c) Compare the two results and explain why there is a...
A bullet of mass 0.017 kg traveling horizontally at a high speed of 210 m/s embeds itself in a block of mass 5 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? Vr = 42 x m/s (b) Calculate the total translational kinetic energy before and after the collision. Ktrans,i = 374.85 Ktrans,f= (c) Compare the two results and explain why there...
A 21.0-kg cannonball is fired from a cannon with muzzle speed of 900 m/s at an angle of 33.0° with the horizontal. A second ball is fired with the same initial speed at an angle of 90.0°. Let y = 0 at the cannon. (a) Use the isolated system model to find the maximum height reached by each ball hfirst ball hsecond ball (b) Use the isolated system model to find the total mechanical energy of the ball-Earth system at...
A cannonball is fired horizontally at 20 m/s from a cliff. Its speed 2 seconds after being fired is about _________ m/s A cannonball is fired horizontally at 20 m/s from a cliff. Its speed 2 seconds after being fired is about m/s Answer in m/s
A 19.0-kg cannonball is fired from a cannon with muzzle speed of 1 050 m/s at an angle of 36.0° with the horizontal. A second ball is fired with the same initial speed at an angle of 90.0°. Let y = 0 at the cannon. (a) Use the isolated system model to find the maximum height reached by each ball. hfirst ball = m hsecond ball = m (b) Use the isolated system model to find the total mechanical energy of the...
In Figure (1), a 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.36 kg) and embeds itself in block 2 (mass 1.87 kg). The blocks end up with speeds v1 = 0.500 m/s and v2 = 1.35 m/s (see Figure (2)). Neglecting the material removed from block 1 by the bullet, find the speed of the bullet as it (a) leaves and (b) enters block 1.