Tag Archives: Kinematics in Two Dimensions

Grantham PHY220 Week 2 Problem 3

A bullet is fired from a gun at a shooting range. The bullet hits the ground after 0.32 seconds. How far did it travel horizontally and vertically in this time if it was fired at a velocity of 1100 m/s?

Solution:

Assuming that the gun was fired horizontally.

Consider the horizontal component of the motion.

v_{o_x}=1100\:m/s

\Delta x=V_{0_x}t=\left(1100\:m/s\right)\left(0.32\:s\right)=352\:m

Consider the vertical component.

v_{o_y}=0\:m/s

\Delta y=V_{0_y}t-\frac{1}{2}gt^2=0-\frac{1}{2}\left(9.80\:m/s^2\right)\left(0.32\:s\right)^2=-0.50

The negative sign of  \Delta y means that the bullet went downward.

Therefore, the bullet traveled 352 m horizontally and 0.50 m vertically downward in 0.32 seconds.

College Physics by Openstax Chapter 2 Problem 22

A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.20×105 m/s2 for 8.10×10-4 s . What is its muzzle velocity (that is, its final velocity)?

Solution:

We are given the following: a=6.20 \times 10^{5} \ \text{m/s}^2; \Delta t=8.10 \times 10^{-4} \ \text{s}; and v_0=0 \text{m/s}.

The muzzle velocity of the bullet is computed as follows:

\begin{align*}
v_f & =v_0+at \\
v_f & = 0\:\text{m/s}+\left(6.20\times 10^5\text{ m/s}^2\right)\left(8.10\times 10^{-4}\:\text{s}\right) \\
v_f & =502\:\text{m/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the muzzle velocity, or final velocity, is 502 m/s. 

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College Physics by Openstax Chapter 2 Problem 21

A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2, and 1.85 ms (1 ms = 10-3 s) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball?

Solution:

We are given the following values: a=-2.10 \times 10^4 \ \text{m/s}^2; t=1.85 \times 10^{-3} \ \text{s}; v_f=0 \ \text{m/s}.

The formula in solving for the initial velocity is

v_0=v_f-at

Substitute the given values

\begin{align*}
v_0 & =0\:\text{m/s}-\left(-2.10\times 10^4\text{ m/s}^2\right)\left(1.85\times 10^{-3}\:\text{s}\right) \\
v_0 & =38.85\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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