Answer with Explanation:
We are given that
Mass of bullet,[tex]m_1=0.03 kg[/tex]
[tex]u_1=470 m/s[/tex]
[tex]m_2=3 kg[/tex]
[tex]\mu_k=0.2[/tex]
[tex]m_3=30 kg[/tex]
We have to find the velocity of the bullet and block B after the first impact and final velocity of the carrier.
According to law of conservation of momentum
[tex]m_1u_1=(m_1+m_2)v[/tex]
[tex]0.03(470)=(0.03+3)v[/tex]
[tex]v=\frac{0.03(470)}{(0.03+3)}=4.65 m/s[/tex]
Hence, the velocity of the bullet and block B after the first impact=4.65 m/s
According to law of conservation of momentum
[tex](m_1+m_2)v=(m_1+m_2+m_3)V[/tex]
[tex](0.03+3)\times 4.65=(0.03+3+30)V[/tex]
[tex]V=\frac{(0.03+3)\times 4.65}{(0.03+3+30)}[/tex]
[tex]V=0.43 m/s[/tex]
Answer:
[tex]v_B=4.65\ m.s^{-1}[/tex] is the velocity of the block and the bullet system after the first impact.
[tex]v_c=0.43\ m.s^{-1}[/tex] is the velocity of the carrier.
Explanation:
Given
mass of bullet, [tex]m=0.03\ kg[/tex]
horizontal velocity of bullet, [tex]v=470\ m.s^{-1}[/tex]
mass of block, [tex]m_B=3\ kg[/tex]
mass of carrier, [tex]m_c=30\ kg[/tex]
coefficient of kinetic friction between block and carrier, [tex]\mu_k=0.2[/tex]
Since the collision between the bullet and the block is perfectly plastic:
We have the conservation of linear momentum as:
[tex]m.v=(m+m_B).v_B=(m+m_B+m_c).v_c[/tex]
[tex]0.03\times 470=(0.03+3)\times v_B=(0.03+3+30)\times v_c[/tex]
[tex]v_B=4.6535\ m.s^{-1}[/tex] is the velocity of the block and the bullet system after collision.
After the collision of block & carrier:
[tex]v_c=0.4268\ m.s^{-1}[/tex] is the velocity of the carrier.
