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4 years ago

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During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 0.148 kg baseball crashing through the pane

of a second-floor window in a nearby building. The ball strikes the glass at 15.5 m/s , shatters the glass as it passes through, and leaves the window at 10.1 m/s with no change of direction. What is the direction of the impulse that the glass imparts to the baseball

1 answer:

LiRa [457]4 years ago

3 0

Answer:

Explanation:

mass of baseball, m = 0.148 kg

initial velocity, u = 15.5 m/s

final velocity, v = 10.1 m/s

Impulse is defined as the change in momentum of the body.

Impulse = change in momentum

I = m (v - u)

I = 0.148 (10.1 - 15.5)

I = - 0.8 Ns

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At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another b

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Answer:

(a) The two balls collide $2\; \rm s$ after launch.

(b) The height of the collision is $4\; \rm m$ .

(Assuming that air resistance is negligible.)

Explanation:

Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.

The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of $g$ should be negative. The question states that the magnitude of $g\!$ is $10\; \rm m \cdot s^{-2}$ . Hence, the signed value of $\! g$ should be $\left(-10\; \rm m \cdot s^{-2}\right)$ .

Similarly, the initial velocity of the ball thrown downwards should also be negative: $\left(-8.0\; \rm m \cdot s^{-1}\right)$ .

On the other hand, the initial velocity of the ball thrown upwards should be positive: $\left(12\; \rm m \cdot s^{-1}\right)$ .

Let $v_0$ and $h_0$ denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time $t$ :

$\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0$ .

For both balls, $g = \left(-10\; \rm m \cdot s^{-2}\right)$ .

For the ball thrown downwards:

Initial velocity: $v_0 = \left(-8.0\; \rm m \cdot s^{-1}\right)$ .
Initial height: $h_0 = 40\; \rm m$ .

$\displaystyle h(t) = -5\, t^{2} + (-8.0)\, t + 40$ (where $h$ is in meters and $t$ is in seconds.)

Similarly, for the ball thrown upwards:

Initial velocity: $v_0 = \left(12\; \rm m \cdot s^{-1}\right)$ .
Initial height: $h_0 = 0\; \rm m$ .

$\displaystyle h(t) = -5\, t^{2} + 12\, t$ (where $h$ is in meters and $t$ is in seconds.)

Equate the two expressions and solve for $t$ :

$-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t$ .

$t = 2$ .

Therefore, the collision takes place $2\, \rm s$ after launch.

Substitute $t = 2$ into either of the two original expressions to find the height of collision:

$h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m$ .

In other words, the two balls collide when their height was $4\; \rm m$ .

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