Question

The cart has a mass of 3 kg and rolls freely from A down the slope. When it reaches the bottom, a spring loaded gun fires a 0.5-kg ball out the back with a horizontal velocity of m/s, measured relative to the cart. Determine the final velocity of the cart.

Answer 1

Answer:

V_c=5.037 m/s

Explanation:

SOLUTION:

applying the conversation of energy, taking the datum at B

$T_{A} +V_{A} =T_{B} +V_{B}$

$T_{A} =\frac{1}{2}(m_{c} +m_{b} )v_{a^2} =\frac{1}{2}(3+0.5)(0)^2\\V_{A}=(m_{c} +m_{b} )gh_{A} =(3+0.5)*9.81*1.25=42.918J\\T_{B}=\frac{1}{2}(m_{c} +m_{b} )v_{B} ^2=\frac{1}{2}(3+0.5)v_{B} ^2=1.754v_{B} ^2$

V_b=(m_c+m_b)gh_a=(3+0.5)*9.81*0=0 J

V_b=4.95 m/s

applying the conversation of linear momentum,

(m_c+m_b)V_b=m_c*V_c+m_b*V_b

V_b=6*V_c-34.66            (1)

utilising the relative velocity relation,

V_b/c=V_b-V_c

V_b=0.6-V_c                      (2)

from (1) (2)

6*V_c-34.66=0.6-V_c

7*V_c=35.26

the final velocity of the cart is

V_c=5.037 m/s

into (2)

V_b=0.6-V_c    

V_b=0.6-5.037

      = 4.437

   

Answer 2

Answer:    v = (3u/3.5)m/s

Explanation:

Initial velocity = u m/s

Final velocity = v m/s

Total mass of cart and bullet = (3 + 0.5) Kg = 3.5 Kg

Momentum before collision = 3 X u m/s = 3u

Momentum after collision = 3.5 X v m/s = 3.5v

∴ Momentum before collision = Momentum after collision

3u = 3.5v

v = (3u/3.5)m/s