Question

Homework H.4.M Given: A cannonball P of mass m is fired toward a steel barrier on a stationary cart. At some time after rebounding from the barrier, the cannonball is observed to have a speed of up and is moving in the direction shown below in the figure. Let M be the combined mass of the cannon/cart. Assume that the cart is able to move without friction along the horizontal surface and ignore the influence of air resistance. Find: For this problem: (a) Determine the velocity vector of the cart after the cannonball bounces off the steel barrier at the instant shown below; (a) If At represents the elapsed time between the firing of the cannonball and the instant shown below, determine the average value of the horizontal force acting on the combined cannon/cart over the time period of 0 <t< At. cannon cart Use the following parameters in your analysis: mg = 80 lb, Mg = 240 lb, At = 0.3 s, 0 = 20° and Up = 100 ft/s.

Answer 1

3 3 solution : The valius provided in the question are : CIM Weight of Cannonball = mg = 80th Weight of cart/ canson = Mg = 240 l velocity of cannonball = F, where I pl-looft's Upwand angle of cannonball velocity =0= 20 Elapsed ticke lutween fining and instant shown- At=0-33 we're assumed that cant is able to move without friction horizontally and 5 no air resistance Hence, we can consider cart / cannon + cannonball system as a closed, inolated system. Sherefore, from the theory of collision we can say that the linear momentum of each colliding body can change, but the net linear momentum will not change. let us say the final velocity of cart / cannon = ve nuu WAN

Also, let us say, impulse in' horizontal directions Je initial momentum of system in horizontal direction - Per mum uy final momentum of system in horizontal direction - Hence, Je = Byt - Pie (Implete lines and then @ Momeuluen tres hem) @ therefore, applying for the enlíne systems De = - Fix = 0 - Also, since initially system was stationary . Biro n was stationary, » from 1 and it , it implies that 1 hel - Fix l-FD

Further, using the given values - Ihrel = m Up caso + Mve - from (4), W and > ming cose + Mivel=0 > vel = -mg Vp cose W = ve = -80 lex 100 H/A * cos 20° 1 240 to solving, we get 've = - 31.3 HA. Hence, the cart will be moving horizontally with 31-3fta velocity but in opposite direction of the Cannon ball. ☺ of the amrage magnitude of the horizontal force in F, we can

white the magnitude of the impelse as J - Fiat - Applying impulse linean momentum theorem © on the cart/ cannon Changers in momentum = Imprese > J = M Ve -(vii Uling D Vi → = 240 l x 31-3 ft/6 (9= 32.17 /18) = 32 0.3 A x 32 = 778.4 le Hence, the querage value of the horizontal force alling on the combined cannon/ cant lover the time period of oct Lat is 778.4 lle.

NOTE:

1. There are four pages uploaded in sequence.

2. Theorem used: Impule-Linear Momentum. For the 1st part of the question (i.e. (a)) this theorem is applied on the entire system of cannon/cart and cannonball, however, for the 2nd part of the question (i.e. (b)) this theorem is applied only on combined cannon/cart.

3. Answer of (a) is that the cart will be moving horizontally with 31.3 ft/s velocity in opposite direction of that of the cannonball's horizontal velocity.

4. Answer of (b) is the average value of the horizontal force acting on the combined cannon/cart over the time period of 0 < t < $\Delta$ t is 778.4 lb.