Problem :
A sphere of mass $3/8kg$. is projected vertically upwards with velocity $7 m/sec$. from a point below the ceiling of the room at a distance $1.6 metre$. It impinges the ceiling, then it rebounded downwards. If the magnitude of change in the magnitude of its momentum due to this impact with the ceiling equals $2400 gm.m./sec$. Calculate the velocity of rebounding of the sphere.
So it should look like this :
My attempt :
to get $V_1$ :
$V_o = 7 m/s$
$a = g = -9 m/sec^2$
$d = 1.6m$
So $V_1^2 = (7)^2 + (2 \times -9.8 \times 1.6)$
$V_1^2 = 17.64$
$V_1 = 4.2m/s$
Since Change of Momentum = $2400 gm.m/sec$
$2400 = ({3 \over 8} \times 10^{3})(V_2 - V_1)$
$2400 = ({3 \over 8} \times 10^{3})(V_2 - 4.2)$
then $V_2 = 10.6m/s$ (my answer)
However my textbook says that the correct answer should be $2.2m/s$
i have no idea what went wrong , any help will be appreciated
thanks in advance.
You're so close! Remember, momentum is a vector, so in most problems involving momentum, direction is important. In your equation $$2400 = \left(\frac{3}{8} \times 10^3\right)(V_{2}-4.2)$$ You have implicitly defined a positive velocity as an upward velocity. Good choice. However, in this case, your change in momentum is negative, so your left hand side of the equation needs a negative sign. You should write $$-2400 = \left(\frac{3}{8} \times 10^3\right)(V_{2}-4.2)$$ You will recover an answer of $-2.2\ m/s$. This is the ball's velocity, and it makes sense that it is negative since the ball is travelling downward after the rebound. Since the problem is really asking about the speed, just take the absolute value to obtain $2.2\ m/s$.