Question

Ballistic Pendulum Version

Procedure: Fire a practice shot. Locate the landing point on the floor then secure a piece of white paper to the floor (centered about the landing point). Cover the sheet with a piece of carbon paper and tape it down. 1. Shoot the ball at least 5 times. Take care not to disturb the position of the launcher during the shots. 2. Measure the vertical height H of the projectile launcher from the floor. (Use a plumb bob to determine the true vertical distance to the floor). Record value in the Table. 3. Measure the horizontal displacement R for each trial and record values in the table. 4. Calculate the average <R> of the horizontal distance. 5. Find the initial velocity vo of the projectile using the <R> and H values. Data Table 1: Trial # Range R(m) 1 2.260 Vertical Height H = 0.925 m Absolute error in distance AH = AR = +0.001 m Horizontal range <R> (m) Time of flight t = (sec) Initial Velocity Vo= (m/s) 2 2.264 3 2.280 4 2.254 5 2.250 <R> Part II: Ballistic Pendulum Method Rem PE xo h KE Page 3 of 7

Answer 1

Date 1 202 Part I: I Table 1 Average sange <R> = 2:260 +2.264+2:280 +2.254 +2:257 5 <R> = 2.2616 (Ans) 2K Hence, Horizontal Range <R>= 2:2616 m Time of flight t = where, His vertical Height = 0.925m Therefore, ta 2x0.925 t = 0.189 seconds (Ans) Q+ (a) :- Initial velocity to Horizontal Ringe <R? Time of flight t Vo = 2.26165 0.001 0.189 Yo = 11.966 ± 0.005 m/s =9.8 ms 9.8

ME 0:06586+ 0.24709_ Part II Total mass (M = maty Mp) 0.24709 Kg me= 0,06586 kg e Me M = 0.31295 kg Length of Pendulum Rest = 0,290 m h = RM (1- cos o 7 h = 0.290(1-coso) for trialt 0 = 35° h = 0.290[1-60535°) - 0.0524 h=0.0524 7 for trial 2, - = 36 h = 0.290(1-c0336) - 0.0.554 h = 0.0554 ind for trial 3, 4 O = 36 Therefore h = 0.0554 for trials 0 =35° h = 0.0524 Therefore Average h = che 0.0524 +0.0554+0.0554+0:055440_052 5 0.0542 <b>

Average E <O> = 35+36736+36135 <O> = 35.6 Initial velocity = ق eviur vertical displacement <h? = 0.0542 Absolute evion in angle do = +4° 2gh. where his average height or <h> I absolute Absolute error in Rom = DRM 10.001 6 h = Rom (1-coso) aug O = (35.6+ 1) Hence error in I-coso d (1-coso). Ao (60). sino - E1). Sin (35.6) (+ 1x Isin (35.6) All-coso) - do 180 = + 0.01

th=a Ront All-roso) Ah = F (0.00) + 0.00 = t o.oll ha 0:0.542+ 0.011

1.6 v dargah 1 2.x 9.8 x 0.0542 v = 1.03 m/s Ab, Crnon in velocity que Aw: 1 10.011) 70.0542)-72 1(Ah) ht 2. 2 Av = 0.0236 Therefore, initial velocity 0 =(1,03+0.0236 ) m/s (Ans)