Carl has a cannon mounted to a small railroad cart that can move with negligible
ID: 1359984 • Letter: C
Question
Carl has a cannon mounted to a small railroad cart that can move with negligible friction along a horizontal track. He intends to propel the cart by firing two cannon balls but he is clumsy and confused about conservation of linear momentum. Therefore please provide a good explanation of conservation of linear momentum as you answer the questions. Carl weighs 80 kg, the cart weighs 30 kg and there are two 10 kg cannon balls on board. Don’t worry about the weight of the gunpowder etc. You can assume the cannon ball always leaves the cannon with a speed of 50 km/hour with respect to the cannon. With the cart initially at rest, Carl fires the first cannon ball so that it leaves the cart with an angle of elevation of 45 degrees.
(a) Determine the recoil speed of the cart after the cannon ball is fired.
(b) As Carl reaches for the second cannon ball it unfortunately rolls slowly off the cart. Determine the speed of the cart after this event.
(c) Determine the recoil speed that Carl would have been able to achieve if he was not so clumsy and knew how to use the conservation of linear momentum to his advantage.
Explanation / Answer
Here ,
let the speed after first time canon ball is fired is v1
a) Using conservation of momentum
(30 + 10 + 80) * v1 = 50 * 10 * cos(45)
solving for v1
v1 = 2.94 km/hr
the recoil speed of cart is 2.94 kr/hr
b)
as the ball will drop with the same speed as the cart .
the speed of canon will be same , = 2.94 km/hr
c)
let the speed of cart is v2
Using conservation of momentum
(30 + 80) * (v2 - 2.94) = 10 * (50 * cos(45) - 2.94)
solving for v2
v2 = 5.88 km/hr
the speed of cart will be 5.88 km/hr