Answer: the answer given below
(a) Explanation: The impulse on an object is given by the change in momentum of the object. Before the collision, the bullet has momentum p1 = mv0 and the brick has momentum p2 = 0, since it is stationary. After the collision, the combined bullet-brick system has momentum p3.
Conservation of momentum requires that the total momentum before the collision is equal to the total momentum after the collision:
p1 + p2 = p3
mv0 + 0 = (m + M)V
where V is the velocity of the combined bullet-brick system after the collision. Solving for V, we get:
V = (mv0) / (m + M)
The impulse on the bullet during the collision is equal to the change in momentum of the bullet:
J_bullet = p3 - p1 = (m + M)V - mv0
Substituting the expression for V we found earlier:
J_bullet = (m + M)(mv0) / (m + M) - mv0 = 0
Therefore, the impulse on the bullet is zero during the collision.
On the other hand, the impulse on the brick during the collision is:
J_brick = p3 - p2 = (m + M)V - 0 = (m + M)(mv0) / (m + M) = mv0
Therefore, the magnitude of the impulse acting on the brick is equal to the initial momentum of the bullet, mv0, and it is in the same direction as the initial velocity of the bullet.
In summary, during the collision of the bullet and the brick, the impulse acting on the bullet is zero, while the impulse acting on the brick is mv0 in the direction of the initial velocity of the bullet.
(b) We can use the principle of conservation of momentum to solve for the velocity of the brick-bullet combination just after the collision. The total momentum of the system (bullet, brick, and Earth) is conserved before and after the collision. Initially, only the bullet has momentum, which is given by p1 = m*v0, and the momentum of the brick and Earth is zero. After the collision, the bullet becomes embedded in the brick, and the combined system of the brick-bullet has momentum p2. Since the momentum of the Earth is negligible compared to that of the bullet and brick, we can treat the system as closed and apply conservation of momentum:
p1 = p2
m*v0 = (M + m)*v
where v is the velocity of the combined system just after the collision.
Solving for v, we get:
v = (m*v0) / (M + m)
Therefore, the magnitude of the velocity of the brick-bullet combination just after the collision is:
|v| = |(m*v0) / (M + m)|
The direction of the velocity is upward, as the system swings up after the collision due to the conservation of momentum.
(c) The initial kinetic energy of the system is the kinetic energy of the bullet just before the collision, which is given by:
KE1 = (1/2)mv0^2
The final kinetic energy of the system is the kinetic energy of the combined brick-bullet system just after the collision, which is given by:
KE2 = (1/2)*(M + m)*v^2
Substituting the expression we found for v:
KE2 = (1/2)(M + m)[(mv0) / (M + m)]^2
KE2 = (1/2)(m*v0^2) / (1 + M/m)
The ratio of the final kinetic energy to the initial kinetic energy is:
KE2 / KE1 = [(1/2)(mv0^2) / (1 + M/m)] / [(1/2)mv0^2]
KE2 / KE1 = 1 / (1 + M/m)
Therefore, the ratio of the final kinetic energy of the brick-bullet combination immediately after the collision to the initial kinetic energy of the brick-bullet combination is:
KE2 / KE1 = 1 / (1 + M/m)
(d)To determine the maximum vertical position reached by the brick-bullet combination, we can use conservation of energy, assuming there is no energy loss due to friction or other dissipative forces. At the maximum height, the kinetic energy of the system is zero, and all the initial kinetic energy has been converted to potential energy due to the height above the initial position.
The initial total energy of the system is the sum of the initial kinetic energy of the bullet and the gravitational potential energy of the brick:
E1 = (1/2)mv0^2 + Mgh1
where h1 is the initial height of the brick above the ground, and g is the acceleration due to gravity.
At the maximum height, the final total energy of the system is the potential energy due to the height above the ground:
E2 = (M + m)gh2
where h2 is the maximum height reached by the brick-bullet combination above the initial position.
Since there is no energy loss, we can set the initial energy equal to the final energy:
E1 = E2
Substituting the expressions for E1 and E2 and solving for h2, we get:
(M + m)gh2 = (1/2)mv0^2 + Mgh1
h2 = [(1/2)mv0^2 + Mgh1] / [(M + m)*g]
Simplifying, we get:
h2 = (1/2)v0^2 / g + h1(M/m) / (1 + M/m)
Therefore, the maximum vertical position above the initial position reached by the brick-bullet combination is:
h2 = (1/2)v0^2 / g + h1(M/m) / (1 + M/m)
Hope this helps :)