Question

A firecracker breaks up into several pieces, one of which has a mass of 200 g and flies off along
the x-axis with a speed of 82.0 m/s. A second piece has a mass of 300 g and flies off along the
y-axis with a speed of 45.0 m/s. What are the magnitude and direction of the total momentum
of these two pieces?
1) 361 kg m/s at 56.3 from the x-axis
2) 93.5 kg m/s at 28.8 from the x-axis
3) 21.2 kg m/s at 39.5 from the x-axis
4) 361 kg m/s at 0.983 from the x-axis
5) 21.2 kg m/s at 56.3 from the x-axis

Answer 1

Explanation:

It is given that,

Mass of first piece, $m_1=200\ g=0.2\ kg$

Speed of first piece, $v_x=82\ m/s$

Mass of second piece, $m_2=300\ g=0.3\ kg$

Speed of first piece, $v_x=45\ m/s$

The momentum along x axis is given by :

$p_x=m_1\times v_x$

$p_x=0.2\times 82=16.4\ kg-m/s$

The momentum along y axis is given by :

$p_y=m_2\times v_y$

$p_y=0.3\times 45=13.5\ kg-m/s$

Let p is the total momentum of these two pieces. Its magnitude is given by :

$p=\sqrt{p_x^2+p_y^2}$

$p=\sqrt{16.4^2+13.5^2}$

p = 21.24 kg-m/s

The direction of total momentum is given by :

$tan\theta=\dfrac{p_y}{p_x}$

$tan\theta=\dfrac{13.5}{16.4}$

$\theta=39.4^{\circ}$

So, the magnitude and direction of the total momentum  of these two pieces are 21.2 kg m/s at 39.5 from the x-axis. Hence, this is the required solution.