# Ap Physics C 1995 Question 1?

Jun 12, 2018

a) Impulse is the area under a force time graph.

$A = \frac{10 \times {10}^{-} 3 + 2 \times {10}^{-} 3 \times 2.0 \times {10}^{3}}{2}$

$A = 12 N s$

b) We can't use conservation of energy or conservation of momentum here because we have no information on what happens after the collision.

However we do have the impulse, or the change in momentum, of the ball. Therefore

$\Delta P = m \Delta v$

$12 = 5 \Delta v$

$\Delta V = 2.4$ m/s.

Since the ball started at rest, after the collision, the speed of the ball will be $2.4$ m/s.

c) Now we can apply conservation of momentum.

$0.5 \left(26\right) = 2.4 \left(5\right) + \left(0.5\right) {v}_{f}$

${v}_{f} = 2$ m/s

Since this quantity is positive the direction will be to the right.

d) The kinetic energy at the beginning was simply

$\frac{1}{2} {m}_{\text{block}} {v}^{2} = \frac{1}{2} \left(0.5\right) {\left(26\right)}^{2} = 169$ joules

At the end, it was

$\frac{1}{2} {m}_{\text{block"v_"final"^2 + 1/2m_"ball"v_"final}}^{2} = \frac{1}{2} \left(0.5\right) {\left(2\right)}^{2} + \frac{1}{2} \left(5\right) {\left(2.4\right)}^{2} = 15.4$ Joules

The difference is the kinetic energy lost. Thus, $153.6$ joules are lost.

e) Recall that $d = \frac{1}{2} a {t}^{2}$. The table is at a height of $1.2$ meters and the only acceleration will be due to gravity. Thus the two objects land at the same time, namely $0.495$ seconds.

We know that $d = v t$. For the block, it will land at $2 \left(0.495\right) = 0.99$ meters from the table. The ball meanwhile will land at $2.4 \left(0.495\right) = 1.19$. Therefore, the ball and the block will land $0.20$ meters apart.

Hopefully this helps!