A ballistic pendulum is used to measure the speed of high-speed
projectiles. A bullet *A* weighing 22 g is
fired into a 2.6-kg wood block *B* suspended by a
cord of length* l* = 2.2 m. The block then swings
through a maximum angle of *θ* = 60°.

a) The initial speed of the bullet is_____m/s

b) The impulse imparted by the bullet on the block is____J

c)The force on the cord immediately after the impact is_____N

Please, explain in details. Thanks!

I would aprecciate the correct answers*

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(a)Apply the law of conservation of energy.\frac{1}{2}\left(m_{b}+m_{B}\right) v_{f}^{2}=\left(m_{b}+m_{B}\right) g hh= the height to which the bullet and block rise after impact.v_{f}=\sqrt{2 g h}The height can be determined as,\begin{aligned}h & =l-l \cos \theta \\& =2.2-2.2 \cos 60^{\circ} \\& =1.1 \mathrm{~m} .\end{aligned}Therefore the final velocity of the bullet is,\begin{aligned}v_{f} & =\sqrt{2 g h} \\v_{f} & =\sqrt{2 \times 9.8 \times 1.1} \\v_{f} & =4.64 \mathrm{~m} / \mathrm{s} .\end{aligned}Apply the conservation of momentum principle.\begin{array}{l}m_{b} v_{0}+m_{B} v_{B}=\left(m_{b}+m_{B}\right) v_{f} \\0.022 v_{0}+0=(0.022+2.6) \times 4.64 \\v_{0}=553.00 \mathrm{~m} / \mathrm{s}\end{array}(b)The impulse imparted by the bullet is determined from the change in moment of the bullet.Impulse imparted by the bullet=change in momentum of the bullet\begin{aligned}F \cdot d t & =m_{b}\left(v_{f}-v_{0}\right) \\& =m_{b}\left(v_{f}-v_{0}\right) \\& =0.022(4.64-553) \\& =-12.06 \mathrm{~N} \cdot \mathrm{s}\end{aligned}Thus the impusle imparted by the bullet on the block is 12.06 \mathrm{~N}.s (c)The force in the cable is due to the centrifugal force of the system which is due to the motion of the system in a curved path and the weight of the system.\begin{aligned}F & =\frac{\left(m_{b}+m_{B}\right) v_{f}^{2}}{l}+\left(m_{b}+m_{B}\right) g \\& =(0.022+2.6) \times \frac{4.64^{2}}{2.2}+(0.022+2.6) \times 9.81 \\& =51.37 \mathrm{~N}\end{aligned} ...