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 1

Answer:

the final velocity of the cart is 5.037m/s

Explanation:

Using the conservation of energy

$T_a + V_a = T_b + V_b$

$T_a = \frac{1}{2} (m_c + m_b)v_a^2$

$T_a= \frac{1}{2} (3 + 0.5)(0)^2$

= 0

$V_a = (m_c + m_b)gh_a$

$V_a = (3 + 0.5) * 9.81 * 1.24$

$= 42.918J$

$T _b = \frac{1}{2} (3 + 0.5)v_b^2 \\\\ = 1.75v_b^2$

$V_b = (3 + 0.5) * 9.81 * 0\\ = 0$

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

Using the conservation of linear momentum

$(m_c + m_b)v_B = m_cv_c + m_bv_b\\(3 + 0.5) * 4.95 = 3v_c - 0.5v_b\\17.33 = 3v_c - 0.5v_b\\v_b = 6v_c - 34.66 ...............(1)$

$Utilizing the relative velocity relation = v_b - v_c\\-0.6 = -v_b - v_c\\v_b = 0.6 - v_c (2)$

equate (1) and (2)

$6v_c - 34.66 = 0.6 - v_c\\7v_c = 35.26\\v_c = 5.037m/s$

the final velocity of the cart is 5.037m/s