A 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.16 kg) and embeds itself in block 2 (mass 1.85 kg). The blocks end up with speeds v1 = 0.560 m/s and v2 = 1.36 m/s (Figure (2)). Neglecting the material removed from block 1 by the bullet, find the speed of the bullet as it (a) enters and (b) leaves block 1.
m =0.0035 kg, m1 =1.16 kg,
m2 = 1.85 kg,
v1 =0.560 m/s, v2 = 1.36 m/s,
u1 =?, u2 =?
When the bullet embeds itself into
block #2, we have a completely inelastic collision. Momentum is conserved and the two objects move together as one. The conservation equation yields:
m•u2 = (m+m2)•v2,
u2 = (m + m2)•v2/m =
= (0.0035+1.85) •1.36/0.0035 =720 m/s,
For the first collision:
m•u1= m•u2 +m1•v1,
u1= (m•u2 +m1•v1)/m =
=(0.0035•720+1.16•0.560)/0.0035 =
= 905.6 m/s,
Thank You!
To find the speed of the bullet as it enters and leaves block 1, we can use the principle of conservation of linear momentum.
The principle of conservation of linear momentum states that the total momentum of a closed system remains constant if no external forces act on it. In this case, the bullet and the two blocks form a closed system since there are no external forces acting on them.
The linear momentum of an object is defined as the product of its mass and velocity:
Momentum = mass x velocity
Before the collision, the bullet has a mass of 3.50 g. Let's convert the mass to kg by dividing it by 1000:
Mass of the bullet, m1 = 3.50 g / 1000 = 0.00350 kg
Let the initial velocity of the bullet be v0 (as it enters block 1) and the final velocity of the bullet be v2 (as it leaves block 1).
Using the principle of conservation of linear momentum, we can write the following equation:
Initial momentum = Final momentum
(mass of bullet x initial velocity of bullet) = (mass of block 1 x velocity of block 1) + (mass of block 2 x velocity of block 2)
(0.00350 kg x v0) = (1.16 kg x 0.560 m/s) + (1.85 kg x 1.36 m/s)
Solving this equation will give us the values of v0, the initial velocity of the bullet as it enters block 1:
0.00350 kg x v0 = (1.16 kg x 0.560 m/s) + (1.85 kg x 1.36 m/s)
v0 = [(1.16 kg x 0.560 m/s) + (1.85 kg x 1.36 m/s)] / 0.00350 kg
v0 ≈ 958.57 m/s
Therefore, the speed of the bullet as it enters block 1 is approximately 958.57 m/s.
To find the speed of the bullet as it leaves block 1, we can use the same principle of conservation of linear momentum. However, this time, we need to consider the mass and velocity of block 1 after the bullet embeds itself in it.
Let the final velocity of block 1 be v1' (after the bullet embeds itself in it).
Applying the principle of conservation of linear momentum:
(mass of bullet x initial velocity of bullet) = (mass of block 1 x final velocity of block 1) + (mass of block 2 x velocity of block 2)
(0.00350 kg x v0) = (1.16 kg x v1') + (1.85 kg x 1.36 m/s)
Solving this equation will give us the value of v1', the final velocity of block 1 after the bullet embeds itself in it. This will also give us the speed of the bullet as it leaves block 1.
v1' = [(0.00350 kg x v0) - (1.85 kg x 1.36 m/s)] / 1.16 kg
v1' ≈ -0.791 m/s
Since the final velocity, v1', is negative, it indicates that block 1 is moving in the opposite direction to the original motion of the bullet. The magnitude of the velocity is 0.791 m/s.
Therefore, the speed of the bullet as it leaves block 1 is approximately 0.791 m/s.