2-A 10.0 Kg block is released from point A in the figure below. The block travels along a smooth frictionless track except for the rough 1 m long portion between points B and C. It makes a collision with stationary 10 kg ball after passing point C. The ball moves forward with speed of 2 m/s.
a) What is the speed of the block at point C?
b) What is the velocity of the block after the collision?
c) Does the block return back to the inclined? If It does how high does it get on the incline?
d) Determine whether the collision was elastic or inelastic
. (Show complete work)
A
1m
BC
The cofficient of friction is missing
To solve this problem, we need to apply the laws of conservation of energy and conservation of momentum. Let's break down the problem step by step:
a) To find the speed of the block at point C, we need to consider the conservation of energy. The block starts at point A with only potential energy and no kinetic energy. As it moves down the incline, potential energy is converted into kinetic energy. The rough portion between B and C will cause some energy loss, but we need to find the speed just before the collision with the ball.
To find the speed at point C, we can use the principle of conservation of mechanical energy, which states that the total mechanical energy (PE + KE) is conserved in the absence of dissipative forces like friction.
The mechanical energy at point A is given by:
(Energy at A) = m*g*h, where m is the mass of the block, g is the acceleration due to gravity, and h is the height of point A.
The mechanical energy at point C is given by:
(Energy at C) = KE + PE, where KE is the kinetic energy and PE is the potential energy at point C.
Since the block is released from rest at point A, the potential energy at point A is equal to the mechanical energy at A.
Therefore, (Energy at A) = (Energy at C)
m*g*h = (1/2)*m*v^2 + m*g*0
where v is the speed at point C.
Simplifying the equation, we get:
10*9.8*1 = (1/2)*10*v^2
98 = 5v^2
v^2 = 98/5
v = √(98/5) ≈ 4.42 m/s
Therefore, the speed of the block at point C is approximately 4.42 m/s.
b) To find the velocity of the block after the collision with the ball, we need to consider the principle of conservation of momentum. This principle states that the total momentum before the collision is equal to the total momentum after the collision if no external forces act on the system.
The momentum before the collision is given by:
(momentum before) = m1*v1, where m1 and v1 are the mass and velocity of the block.
The momentum after the collision is given by:
(momentum after) = m1*v + m2*v2, where m2 is the mass of the ball and v2 is the velocity of the ball.
Since the ball is stationary before the collision (v2 = 0), the equation becomes:
(momentum before) = (momentum after)
10*4.42 = 10*v + 10*2
44.2 = 10v + 20
10v = 44.2 - 20
10v = 24.2
v = 24.2/10 ≈ 2.42 m/s
Therefore, the velocity of the block after the collision with the ball is approximately 2.42 m/s.
c) To determine if the block returns back to the incline, we need to consider the conservation of energy again. After the block collides with the ball, it will continue moving with a certain velocity. Let's assume it reaches point D on the incline before coming to a stop.
The mechanical energy at point C is given by:
(Energy at C) = (1/2)*m*v^2 + m*g*h', where h' is the height of point D.
Since the block comes to a stop at point D, the kinetic energy at point D is zero.
Therefore, (Energy at C) = m*g*h'
Using the same numerical values as before, we can solve for h':
98 = 10*9.8*h'
h' = 98/(10*9.8) ≈ 1 m
Therefore, the block does return back to the inclined plane, and it reaches a height of approximately 1 m on the incline.
d) To determine whether the collision was elastic or inelastic, we need to analyze the kinetic energy before and after the collision. If the kinetic energy is conserved, it is an elastic collision. If the kinetic energy decreases, it is an inelastic collision.
Before the collision, the kinetic energy of the block is given by:
(KE before) = (1/2)*m*v1^2, where v1 is the velocity of the block before the collision.
After the collision, the kinetic energy of the block is given by:
(KE after) = (1/2)*m*v^2, where v is the velocity of the block after the collision.
Calculating the kinetic energies:
(KE before) = (1/2)*10*4.42^2 ≈ 97.958 J
(KE after) = (1/2)*10*2.42^2 ≈ 29.268 J
Since the kinetic energy after the collision is less than the kinetic energy before the collision, the collision is inelastic.
Therefore, the collision between the block and the ball is an inelastic collision.