Question

Part A A small spring-loaded launcher is designed to fire steel ball bearings with masses of 0.28 kg. When setting the launcher, you compress the spring a distance of 13 cm. You then fire the launcher straight up and find that the projectiles reach a maximum height of 1 m above the point that they leave the launcher. You load another ball bearing, compress the spring a distance of 19 cm, and place it horizontally at the edge of a table that is 0.5 m tall. You fire the ball from the edge of the table. How far from the base of the table does it hit the ground? PO AQ R O 2 ? 1.76 Submit Previous Answers Request Answer X Incorrect; Try Again; 6 attempts remaining < Return to Assignment Provide Feedback

Answer 1

The potential energy of spring when compressed by amount $\Delta x$ is

$U_{spring}=\dfrac{1}{2} k\Delta x^2$

where k is the spring constant.

The gravitational potential energy of the ball at height h is

$U=mgh$

Given that when the spring was compressed by amount 13cm=0.13m, it pushes the ball to the height of 1m above the point of release.

$\dfrac{1}{2} k\Delta x^2=mgh$

$k=\frac{2mgh}{\Delta x^2}$

$k=\frac{2\times (0.28\ kg)\times (9.81\ m/s^2)\times (1\ m)}{(0.13\ m)^2}$

$k=325.1\ N/m$

Now, the spring is compressed by amount 19cm=0.19m. The kinetic energy that the spring can impart to the ball is

$\dfrac{1}{2} mv^2=\dfrac{1}{2} k \Delta x^2$

$v=\sqrt{\dfrac{k\Delta x^2}{m}}$

$v=\sqrt{\dfrac{(325.1\ N/m)\times (0.19\ m)^2}{(0.28\ kg)}}$

$v=6.474\ m/s$

The ball is launched at speed 6.474m/s horizontally. The ball will undergo a projectile motion with an initial velocity of 6.474m/s in the horizontal direction. In projectile motion, the vertical motion is a uniformly accelerated motion governed by the kinematic's equations for uniformly accelerated motion. While the horizontal motion is constant velocity motion.

The initial vertical velocity of the ball is zero. The ball traveled 0.5m below the point of release before hitting the ground. To find the time t in which the ball reaches the ground from the point of release we use

$h=v_{initial}t&plus;\dfrac{1}{2} at^2$

$-0.5=0\times t&plus;\dfrac{1}{2}\times (-9.81)t^2$

Note that we take height as negative because it is below the ground and acceleration as negative because it points downwards.

$-0.5=-\dfrac{1}{2}\times (9.81)t^2$

$t=\sqrt{\frac{2\times 0.5}{9.81}}$

$t=0.3193\ s$

The ball reaches the ground in 0.3193 seconds. The horizontal distance covered by the ball in this time is

$distance=speed\times time$

$d=(6.474\ m/s)\times (0.3193\ s)$

$\boxed{d=2.07\ m}$