Question a cart mass 3kg rolls down a slope. when it reaches the bottom a spring loaded gun fires a 0.5kg ball with horizontal velocity 0.6m/s. find final velocity of the cart

Answer: the final velocity of the cart is 5.037m/s Explanation: Using the conservation of energy [tex]T_a + V_a = T_b + V_b[/tex] [tex]T_a = \frac{1}{2} (m_c + m_b)v_a^2[/tex] [tex]T_a= \frac{1}{2} (3 + 0.5)(0)^2[/tex] = 0 [tex]V_a = (m_c + m_b)gh_a[/tex] [tex]V_a = (3 + 0.5) * 9.81 * 1.24[/tex] [tex]= 42.918J[/tex] [tex]T _b = \frac{1}{2} (3 + 0.5)v_b^2 \\\\ = 1.75v_b^2[/tex] [tex]V_b = (3 + 0.5) * 9.81 * 0\\ = 0[/tex] [tex]T_a + V_a = T_b + V_b\\ 0 + 12.918 = 1.75v_b^2 + 0\\v_b = 4.95m/s[/tex] Using the conservation of linear momentum [tex](m_c + m_b)v_B = m_cv_c + m_bv_b\\(3 + 0.5) * 4.95 = 3v_c – 0.5v_b\\17.33 = 3v_c – 0.5v_b\\v_b = 6v_c – 34.66 ……………(1)[/tex] [tex]Utilizing the relative velocity relation = v_b – v_c\\-0.6 = -v_b – v_c\\v_b = 0.6 – v_c (2)[/tex] equate (1) and (2) [tex]6v_c – 34.66 = 0.6 – v_c\\7v_c = 35.26\\v_c = 5.037m/s[/tex] the final velocity of the cart is 5.037m/s Log in to Reply

Answer:the final velocity of the cart is 5.037m/s

Explanation:Using the conservation of energy

[tex]T_a + V_a = T_b + V_b[/tex]

[tex]T_a = \frac{1}{2} (m_c + m_b)v_a^2[/tex]

[tex]T_a= \frac{1}{2} (3 + 0.5)(0)^2[/tex]

= 0

[tex]V_a = (m_c + m_b)gh_a[/tex]

[tex]V_a = (3 + 0.5) * 9.81 * 1.24[/tex]

[tex]= 42.918J[/tex]

[tex]T _b = \frac{1}{2} (3 + 0.5)v_b^2 \\\\ = 1.75v_b^2[/tex]

[tex]V_b = (3 + 0.5) * 9.81 * 0\\ = 0[/tex]

[tex]T_a + V_a = T_b + V_b\\ 0 + 12.918 = 1.75v_b^2 + 0\\v_b = 4.95m/s[/tex]

Using the conservation of linear momentum

[tex](m_c + m_b)v_B = m_cv_c + m_bv_b\\(3 + 0.5) * 4.95 = 3v_c – 0.5v_b\\17.33 = 3v_c – 0.5v_b\\v_b = 6v_c – 34.66 ……………(1)[/tex]

[tex]Utilizing the relative velocity relation = v_b – v_c\\-0.6 = -v_b – v_c\\v_b = 0.6 – v_c (2)[/tex]

equate (1) and (2)

[tex]6v_c – 34.66 = 0.6 – v_c\\7v_c = 35.26\\v_c = 5.037m/s[/tex]

the final velocity of the cart is 5.037m/s