Question

An object of mass 10 kg is launched from from the top of building at t=0 , at an angle of 37 degrees above the horizontal with a speed of 30 m/s . After some time , an explosion splits the projectile into two pieces . one piece of mass 4 kg is observed at ( 105 , 45 ) meters at t=2 secs . find the location of second piece at t=2 secs ----

Answer 1

Answer:

t=2seconds

m=10kg=10000g

speed=30m/s

Answer 2

Answer:

The location of second piece at 2 sec is (6.76, 12.3)

Explanation:

Given that,

Mass of object M= 10 kg

Angle = 37°

Speed = 30 m/s

Mass of one piece m= 4 kg

Time = 2 sec

Height = 4.5 m

Distance = 10.5 m

We need to calculate the mass of second piece

$m'=M-m$

$m'=10-4=6 kg$

Position of Center of mass will not change,

So,  at 2 s, Position of the original mass is

We need to calculate the horizontal position of center of mass

Using formula of distance

$s_{x}=u\cos\theta\times t$

Put the value into the formula

$s_{x}=30\cos37\times2$

$s_{x}=8.26\ m$

We need to calculate the vertical position of center of mass

Using formula of distance

$s_{y}=u\sin\theta\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$s_{y}=30\sin37\times2-\dfrac{1}{2}\times9.8\times4$

$s_{y}=9.23\ m$

The point of center of mass is (8.26, 9.23).

We need to calculate the location of second piece

Using formula of center of mass

In horizontal,

$s_{x}=\dfrac{mx_{1}+m'x_{2}}{m+m'}$

Put the value into the formula

$8.26=\dfrac{4\times105+6\times x}{10}$

$x=\dfrac{8.26\times10-4\times10.5}{6}$

$x=6.76\ m$

In vertical,

$s_{y}=\dfrac{mx_{1}+m'x_{2}}{m+m'}$

Put the value into the formula

$9.23=\dfrac{4\times45+6\times y}{10}$

$y=\dfrac{9.23\times10-4\times4.5}{6}$

$y=12.3\ m$

Hence, The location of second piece at 2 sec is (6.76, 12.3).