A 0,1kg bullet is fired horizontally towards blcok A and B, which are hanging from cords. The bullet passes through block A of mass 2kg and becomes embedded in block B of mass 6.4kg. The bullet thus causes A and B to start moving with velocities of 6m,s- and 2m.s- respectively, imeediately after impact. Calculate:
a) the initial velocity of the bullet
b) the velocity of the bullet as it travels from A to B.
Assume u = Initial velocity of the bullet and v= its vel after leaving block A.
V = Vel. of block A after the impact = 2 met/sec
M = Mass of block A = 2 kg & m = mass of the bullet = 0.1 kg
By the principle of conservation of momentum: m*u = m*v + M*V
Or 0.1u = 0.1v + 2*6
Applying the same principle to block B: 0.1v = (6.4 +0.1*2
which gives us: v =130 met/sec
Plug this value in the earlier equation to obtain: u = 250 met/sec
Thus, initial bullet vel. =250 met/sec and Buleet vel. as travels from A to B = 130 met/sec
To solve this problem, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
a) Let's consider the initial situation before the bullet hits block A. We know the mass of block A (2 kg) and the initial velocity of block B (0 m/s) since it is initially at rest. Also, the bullet has a mass of 0.1 kg, and we need to find its initial velocity.
The total momentum before the collision is zero because there is no initial motion. So, the total momentum after the collision should also be zero. We can express this as:
(0.1 kg)(V) + (2 kg)(0 m/s) = 0, where V is the initial velocity of the bullet.
We can solve this equation to find the initial velocity of the bullet:
0.1V = 0
V = 0 m/s
Therefore, the initial velocity of the bullet is 0 m/s.
b) Now, let's consider the situation after the bullet has become embedded in block B. We know the masses of blocks A (2 kg) and B (6.4 kg) and the velocities of A (6 m/s) and B (2 m/s). We need to find the velocity of the bullet as it travels from A to B.
Using the conservation of momentum principle again, we can write:
(mA)(vA) + (mB)(vB) = (mA + mB)(V), where mA and mB are the masses of blocks A and B, vA and vB are their velocities, and V is the velocity of the bullet.
Substituting the given values, we get:
(2 kg)(6 m/s) + (6.4 kg)(2 m/s) = (2 kg + 6.4 kg)(V)
Simplifying this equation:
12 kg·m/s + 12.8 kg·m/s = 8.4 kg( V )
24.8 kg·m/s = 8.4 kg( V )
V = 24.8 kg·m/s / 8.4 kg
V ≈ 2.95 m/s
Therefore, the velocity of the bullet as it travels from A to B is approximately 2.95 m/s.