Question

In an intense battle, gunfire can so concentrated that bullets from opposite sides collide in midair. Suppose that a bullet (with mass M=5.12 g moving to the right at a speed V=202 m/s directed 21.3 ∘ above the horizontal) collides and fuses with another with mass m=3.05 g moving to the left at a speed v=282 m/s directed 15.4∘ above the horizontal. What is the magnitude of their common velocity immediately after the collision?

Answer 1

To determine the magnitude of the common velocity of two bullets colliding and fusing together, we use the conservation of momentum, resolve velocities into components, and apply the Pythagorean theorem to find the resultant velocity.

Step-by-step explanation:

The question involves an inelastic collision in which two bullets collide and fuse together. To find the magnitude of their common velocity immediately after the collision, we can use the conservation of momentum. The total momentum before the collision must equal the total momentum after the collision because no external forces are acting on the bullets.

Let's denote M as the mass of the first bullet 5.12 g (or 0.00512 kg), V as its velocity 202 m/s at 21.3° above the horizontal, m as the mass of the second bullet 3.05 g (or 0.00305 kg), and v as its velocity 282 m/s at 15.4° above the horizontal. We can resolve the velocities into horizontal and vertical components and apply the conservation of momentum for each direction separately:

• Momentum in the horizontal direction: M*V*cos(21.3°) + (-m*v*cos(15.4°)) = (M+m)*V'x
• Momentum in the vertical direction: M*V*sin(21.3°) + m*v*sin(15.4°) = (M+m)*V'y

Here, V'x and V'y are the components of the common velocity after the collision, which can be found using the given masses and velocities. Finally, the magnitude of the common velocity (V') can be determined by using the Pythagorean theorem:

V' = √(V'x^2 + V'y^2)