Recent content by simon1987

Hmm...kilometres per year is an interesting unit.

Looks good.

Does it match what you know the answer should be?

Do you not understand the concept? Or how to solve the equations for x3 and y3, respectively? Or what x3 and y3 represent? Or what CMx and CMy represent?

Solve the equations for x3 and y3 before plugging in numbers.

It's much easier than you are making it. The equation CM_{x} = \frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3}}{m_{1} + m_{2} + m_{3}} Will give you the x-coordinate for the position of the center of mass: CM_{x}. You are looking for the position of the 4-kg mass, so solve this equation...

Bingo!

Have you drawn a picture? It is incredibly helpful when considering multiple different vectors. Consider: what would the cross product be of z and P if z were parallel to P? As for part (2), remember that "combined" is just the sum of the two angular momenta, and that each angular momentum is...

Throughout the equation you were (correctly) calculating from the point of view of the ball (using the ball's mass, velocities, etc.). The question asks what the average force exerted by the club onto the ball is. Re-read Newton's laws for a hint.

There are very simple equations one can use to solve these.

To answer part (1), consider this: In rotational motion, angular momentum \stackrel{\rightarrow}{L} = \stackrel{\rightarrow}{r} \times\stackrel{\rightarrow}{P}, where P is linear momentum.

List everything you know: m1, x1, y1 m2, x2, y2 CM(x), CM(y) m3 You need to find x3 and y3. So set up the equation to find the center of mass, CM{}_x{} = (m{}_1{}x{}_1{} + m{}_2{}x{}_2{} + m{}_3{}x{}_3{}) / (M) Solve it for what you're looking for. EDIT: To be clearer, a...

Then use (m1y1+m2y2+m3y3)/(m1+m2+m3) to find the y-coordinate of the CM.