10 Two balls X and Y are moving towards each other with speeds of 5 m s–1 and 15 m s –1respectively.X5 m s–1Y15 m s–1They make a perfectly elastic head-on collision and ball Y moves to the right with a speedof 7 m s –1.What is the speed and direction of ball X after the collision?A 3 m s –1 to the leftB 13 m s –1 to the leftC 3 m s –1 to the rightD 13 m s –1 to the right11 Two forces form a couple.Which statement describes the two forces?A They are in the same direction.B They are perpendicular to each other.C They have the same magnitude.D They pass through the same point

The question seems to be a bit jumbled, but I'll try to answer each part separately.

For the first question about the two balls, we can use the principle of conservation of momentum. The total momentum before the collision is the same as the total momentum after the collision.

Before the collision, the total momentum is (mass of X * velocity of X) + (mass of Y * velocity of Y). Let's assume the mass of each ball is 1 kg for simplicity. So, the total momentum before the collision is (1 kg * 5 m/s) + (1 kg * -15 m/s) = -10 kg*m/s. (The velocity of Y is negative because it's moving in the opposite direction.)

After the collision, the total momentum is (mass of X * velocity of X) + (mass of Y * velocity of Y). We know the velocity of Y is 7 m/s to the right, so the total momentum after the collision is (1 kg * velocity of X) + (1 kg * 7 m/s).

Setting these two equal to each other, we get -10 kgm/s = (1 kg * velocity of X) + 7 kgm/s. Solving for the velocity of X, we get -17 m/s. So, the speed of ball X after the collision is 17 m/s to the left. This isn't one of the options, so there may be a mistake in the question or my interpretation of it.

For the second question, two forces form a couple if they have the same magnitude, are parallel and opposite in direction, and act along different lines. So, the correct answer is C. They have the same magnitude.