A projectile launcher is set on a table so that the ball becomes a projectile at a height of 1.2 m above the floor. The mass of the ball is 0.01 kg. A plunger is used to push the ball into the barrel of the launcher compressing a spring a distance of...?

0.1 m before it is ready to launch. It is launched at an angle of 30° above the horizontal. The ball travels a horizontal distance of 4 m before striking the floor below. Determine the force constant of the spring that shot the ball in this fashion.

1 Answer
A08
Mar 3, 2018

Let the initial velocity of the projectile be #=u#. Let the origin of coordinates be located at the point of projection.

Horizontal motion.
Distance traveled #=4\ m# during time of flight #t#, with initial horizontal velocity #=ucos30^@#

#:.ucos30^@t=4#
#=>sqrt3/2ut=4# .......(1)

Vertical motion.
To calculate the time of flight we use the kinematic expression

#s=ut+1/2at^2#

Taking #g=9.8\ ms^-2#, noting that gravity acts in the downwards direction and inserting given values we get

#-1.2=usin30^@t+1/2(-9.8)t^2#
#=>4.9t^2-0.5ut-1.2=0# ........(2)

Eliminating #ut# from (2) with the help of (1) we get
#=>4.9t^2-0.5xx8/sqrt3-1.2=0#
#=>t=+-0.846\ s# ........(3)

Ignoring the #-ve# root as time can not be negative. From (1) we get the value of #u# as

#u=8/(sqrt3xx0.846)=5.46\ ms^-1# ......(4)

Now the kinetic energy of the projectile at the time of projection is provided by the potential energy of the compressed spring. We know that

#KE_"projectile"=1/2m u^2# and
#PE_"spring"=1/2kx^2#
where #m# is mass of the projectile, #k# is the spring constant and #x# is the compression of the spring.

Equating both we get

#1/2kx^2=1/2m u^2#
#=>k=m u^2/x^2#

Inserting various values we get

#=>k=0.01 u^2/(0.1)^2#
#=>k=u^2#

Using (4)

#k=29.8\ Nm^-1#

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