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 = (3u/3.5)m/s

Step-by-step 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

Answer 2

V_c=5.037 m/s

Step-by-step explanation:

SOLUTION:

applying the conversation of energy, taking the datum at B

`T_(A) +V_(A) =T_(B) +V_(B)`

`T_(A) =(1)/(2)(m_(c) +m_(b) )v_(a^2) =(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)=(1)/(2)(m_(c) +m_(b) )v_(B) ^2=(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