Momentum, energy?

Question

1) A small explosive charge is placed in a rubber block resting on a smooth surface. When the charge is detonated, the block breaks into 3 pieces. A 200g piece travels at 1.4m/s and a 300g piece travels at 0.90m/s. The 3rd piece flies off at 1.8m/s. If the angle between the first 2 pieces is 80degrees, calculate the mass of the 3rd piece.

2) A 0.40kg cue ball makes a glancing blow to a stationary 0.30kg billiard ball so that the cue ball deflects with a speed of 1.2m/s at an angle of 30degrees from its original path. Calculate the original speed of the cue ball if the billiard ends up travelling at 1.5m/s.

3) A 2kg ball is attached to the end of a string 2.5m long to form a pendulum that is released when the pendulum makes an 80degree angle from the vertical. At the bottom of its swing, the ball collides with a 3kg block initially at rest on a horizontal frictionless surface.If the collision is totally inelastic, what is the maximum height that the masses rise to after the collision?

Answer 1

1. The initial impulse is 0. If all three particles flew in the same direction, you would simply have:

m1 * v1 + m2 * v2 + m3 * v3 = 0

However, as they make an angle with each other, they don't travel in the same direction. You must decompose all speeds in X and Y directions.

m1 * v1x + m2 * v2x + m3 * v3x = 0

m1 * v1y + m2 * v2y + m3 * v3y = 0

The 'good thing' is that you can take the X and Y axes as you wish. So say that v1 is in the direction of Ox. :D

m1 * v1 + m2 * v2x + m3 * v3x = 0

m2 * v2y + m3 * v3y = 0

Now then, you know the angle between the v1 and v2 vectors (80 degrees), and as v1 is in the direction of Ox you have:

v2y = v2 * cos(80º)

Finally, you have the equations:

v2 = sqrt(v2x^2 + v2y^2)

v3 = sqrt(v3x^2 + v3y^2)

You have 5 equations and 5 unknown values (v2x, v3x, v2y, v3y, m3), so you can work the results out. :)

Maybe there's an easier solution, I dunno.

Pay attention to the signs. There is a 80º angle between v1 and v2, so if you take v1 as positive, then v2x must be positive as well, and v3x is negative; furthermore v2y and v3y will have opposite signs.

2. Here too, you can decompose speeds in x and y directions (take the X axis in the direction of the initial velocity vector of the cueball) and work with them until you get the result.

If it is a perfectly elastic collision, you can simply use the conservation of energy.

3. You can calculate the velocity of the ball (right before the collision) from the conservation of energy:

m * v^2 / 2 = m * g * h

where h = 2.5 * (1 - cos80º)

Momentum conservation during the collision:

m * v = (m + M) * u

m = 2, M = 3 kg

The new speed is u. Apply the conservation of energy again to obtain the new height:

(m + M) * u^2 / 2 = (m + M) * g * h1

Result: h1 = (m / (M+m))^2 * 2.5 * (1 - cos80º)