=========================preview======================
(PHYS111)[2010](f)midterm~=tkk933^_29822.pdf
Back to PHYS111 Login to download
======================================================
Physics 111 (Fall 2010) Midterm Solution
1) (a) Time that the ball travels in the air L 13
t.. . 1.759 s . 1.76 s (answer)
v0 cos. 11.5cos50o Remark: Some students calculated the time taken by the ball to reach the height of the basket at 0.3 m, but this is incorrect because the ball does not hit the basket.
(b) Vertical distance
12 o12
y . v0 sin.t . gt . (11.5sin 50 )(1.759) . (9.8)(1.759) . 0.3379 m . 0.3 m
22 The ball is above the basket. Distance from the basket d .0.3379 . 0.3 . 0.0379 m (answer) Remark: Some students used the rounded-up time of 1.76 s to calculate the vertical distance and got 0.0326 m. This is not accurate.
(c) Horizontal velocity
o .1
vx . v0 cos.. 11.5cos50 . 7.392 ms Vertical velocity
o .1
vy . v0 sin.. gt . 11.5sin 50 . (9.8)(1.759) ..8.425 ms
Magnitude
22
2 2 .1 .1
v .
vx . vy . 7.392 . (.8.425) . 11.21ms . 11.2 ms
Direction
.1 vy .1..8.425 . oo
.. tan . tan . ...48.74 ..48.7
vx . 7.392 . The velocity makes an angle of 48.7o with the horizon in the downward direction. (answer)
(d) Using the impulse-momentum theorem, the force acting on the ball is given by mv (0.624)(11.5)
F. 0 .. 71.76 N . 71.8 N (answer)
t 0.1 Remark: This force is different from the force exerted by Yao Ming, because the force is the vector sum of the force exerted by Yao Ming and the weight of the ball. To calculate the force exerted by Yao Ming, we use the impulse-momentum theorem,
mv0 cos. (0.624)(7.392)
x direction: F.. . 46.13 N
xt 0.1
mv0 sin.

y direction: Fy . mg .
t
(0.624)(11.5sin 50o )
. F . (0.624)(9.8) .. 61.09 N
y0.1
22
22
Total force: F .
Fx . Fy . 46.13 . 61.09 . 76.546 N . 76.5 N
Note that the force exerted by Yao Ming is greater than the force acting on the ball due to the weight of the ball.
2) (a) Using the conservation of energy, 12
mgh . mg2R . mv
2
.1 .1
v.
2g(h . 2R) .
(2)(9.8)(0.9 . 0.6) . 2.4249 ms . 2.42 ms (answer)
(b) Let N = normal reaction. Using Newtons law of motion when the ball reaches the top of the loop,
2
v
Radial direction: N . mg . m
R
v22mg . 2h .

N . m . mg . (h . 2R) . mg . mg.. 5.
RR . R .
. (2)(0.9) .

.(0.1)(9.8).. 5.. 0.98 N (answer)
. 0.3 . Remark: Some students used the rounded-up velocity of 2.42 ms.1 to calculate the normal reaction and obtained the value of 0.972 N, but this is not accurate.
(c) Using Newtons law of motion when . = 60o, 2
2 v
v
Radial direction: N . mg cos.. m
R When the ball starts to lose contact with the loop, N = 0. Then
v2 . gRcos60o . 1 gR
2 Using the conservation of energy, 12
mgh . mg(R . R cos. ) . mv
2
o1
mgh . mg(R . R cos60 ) . mgR
4 1 . 11 .
h. (R . Rcos60o ) . R ..1...R . (1.75)(0.3) . 0.525 m (answer)
4 . 24 .
3) (a)-(b) Using the conservation of momentum,
mu . mv1 . 3mv2
u .v1 . 3v2 (1) Using the conservation of energy, 1 21 23 21 . 12 .
mu . mv . mv .. mu .

2 212 28 . 2 .
722 2
u .v1 . 3v2 (2)
8 From (1), v1 . u . 3v2 Subst