SPRING 2002 MIDTERM EXAMINATION 1

PHYSICS 220A #85077

1. A car A at rest passed by car B traveling at a constant 120km/h, chases car B at a constant acceleration. Two cars reach the same destination in 800m. (a) Calculate how long it took car A to overtake car B? (b) What is the acceleration of car A? (c) What is the speed of car A at the end of the chase?

Solution.

(a) Two cars take the same time to reach the same destination, we find the time from car B:

                        x = x0 + vt;

                        800m = 0 + (120km/h)(1000m/km)(1h/3.6ks)t, which gives t = 24s.

(b) We find the acceleration for car A from

                        x = x0 + v0t + 1/2at2;

                        800m = 0 + 0 + 1/2a(24s)2, which gives a = 2.78m/s2.

(c) We find the speed of car A at the end of the chase from

                       v = v0 + at = 0 + (2.78m/s2)(24s) = 66.72m/s.

               

2. A projectile is shot from the edge of a cliff 125m above ground level with an initial speed of 65.0m/s at an angle of 37.0° with the horizontal, as shown. (a) How long does the projectile stay in the air? (b) What is the range of the projectile as measured from the base of the cliff? (c) What is the velocity of the projectile when it hits the ground?

Solution.

 

(a) We choose a coordinate system with the origin aat the base of the cliff as shown.

We find the time required for the fall from the  vertical motion:

                  y = y + v0yt +1/2ayt2;

                  0 = 125 + (65.0m/s)(sin37.0°)t + 1/2(-9.80m/s2)t2,

which gives t = -2.45, 10.4s.

Because the projectile starts at t = 0, we have t = 10.4s.

(b) We find the range from the horizontal motion:

                   x = v0xt = (65.0m/s)(cos37.0°)(10.4s) = 540m.

(c) We find the velocity components from

                   vx = v0x = (65.0m/s)cos37.0° = 51.9m/s.

                   vy = v0y + ayt = (65.0m/s)sin37.0° + (-9.80m/s2)(10.4s) = -62.8m/s.

We find the velocity from the components:

                    v = (vx2 + vy 2)1/2

                       = [(51.9m/s)2 + (-62.8m/s)2]1/2

                       = 81.5m/s.

We find the direction of the velocity from

                    tan q = vy/vx = (-62.8m/s)/(51.9m/s) = -1.21,

which gives q =  -50.4°   (below the horizontal).

3. Two blocks of masses m1 = 5.0kg and m2 = 2.0kg are connected by a light cord, which passes a massless and frictionless pulley as shown. The coefficient of kinetic friction between block m1 and incline is 0.17. The block m1 slide down the incline from rest 9.5m from the base of the incline. Find the speed of the block when it reaches the base.

Solution.

 

 We draw the free-body diagrams for m1 and m2 as shown. We choose the coordinate systems as shown.

We write SF = ma for m1:

                 x-component: m1gsinq - FT - Ffr = m1a;

                 y-component:  FN - m1gcosq = 0.

We have Ffr = mkFN = mkm1gcosq .

We write  for m2:

                y-component: FT  - m2g = m2a.

Combine the equations we have

                  m1gsinq  - m2g - mkm1gcosq = (m1 + m2)a;

                  (5.0kg)(9.80m/s2)sin40° - (2.0kg)(9.80m/s2) - (0.17)(5.0kg)(9.80m/s2)cos40°

                   = (5.0kg + 2.0kg)a, which gives a = 0.79m/s2.    

we find the velocity of m1 when it reaches the base from:     

                    v2 = v02 + 2a(x - x0)

                         = 0 + 2(0,79m/s2)(9.5m - 0), 

which gives v = 3.87m/s.    

4. A 0.5-kg ball suspended by a 2-m cord revolves in a horizontal circle or radius r = 1m. If the period of the circular motion is 2.0s, what is the tension in the cord?

Solution.

        

We find the speed of the circular motion from

                v = 2pr/T = 2p(1m)/(2.0s) =3.14m/s

We find the centripetal force exerted on the ball from

                FC = mv2/r 

                     = (0.5kg)(3.14m/s)2/(1m)

                     = 4.9N.

We draw the free-body diagram for the ball as shown. The x-component of tension in the cord serves as the centripetal force, we have

                 FTsinq = FC ;

                 FT(r/L) = FC;

                  FT(1m/2m) = 4.5N, which gives FT = 9.8N.