A 0.03-lb bullet traveling at 1300 ft/s strikes the 10-lb wooden block and exits the other side at 50 ft/s as shown. Determine the speed of the block just after the bullet exits the block. Also, determine the average normal force on the block if the bullet passes through it in 1 ms, and the time

Question

A 0.03-lb bullet traveling at 1300 ft/s strikes the 10-lb wooden block and exits the other side at 50 ft/s as shown. Determine the speed of the block just after the bullet exits the block. Also, determine the average normal force on the block if the bullet passes through it in 1 ms, and the time the block slides before it stops. The coefficient of kinetic friction between the block and the surface is { \mu }_{ k } = 0.5.

Accepted Answer

[latex]\begin{aligned} (\underrightarrow{ + }) \quad\quad & \Sigma m_1 v_1 = \Sigma m_2 v_2 \ & (\frac { 0.03 } { 32.2 })(1300)(\frac { 12 } { 13 }) + 0 = (\frac { 10 } { 32.2 })v_B + (\frac { 0.03 } { 32.2 })(50)(\frac 4 5) \ & v_B = 3.48 ft/s \end{aligned} \ \begin{aligned} (+ \uparrow) \quad\quad & mv_1 + \Sigma \int F\space dt = mv_2 \ & -(\frac { 0.03 } { 32.2 })(1300)(\frac 5 { 13 }) - 10(1)(10^{ -3 }) + N(1)(10^{ -3 }) = (\frac { 0.03 } { 32.2 })(50)(\frac 3 5) \ & N = 504 lb \end{aligned} \ \begin{aligned} (\underrightarrow{ + }) \quad\quad & mv_1 + \Sigma \int F\space dt = mv_2 \ & (\frac { 10 } { 32.2 })(3.48) - 5(t) = 0 \ & t = 0.216 s \end{aligned}[/latex]

Question : A 0.03-lb bullet traveling at 1300 ft/s strikes the 10-lb wo...

This Question has been Answered.

Already have an account?

Login