Question 1

A block of mass m = 9.0 kg and speed V and is behind a block of mass M = 27 kg and speed of 
As shown int the figure.The surface is frictionless,and the blocks suddenly collide and couple.After the collision,the blocks have a common speed of 0.90 m/s.How much kinetic energy of the blocks is lost due to the collision? 
A)8.6 J
B)2.0 J
C)4.6 J
D)11 J
E)31 J

Accepted Answer

The answer of A block of mass m = 9.0...

Question 2

A 50-g ball moving at 10 m/s in the +x direction suddenly collides head-on with a stationary ball of mass 100 g.If the collision is perfectly elastic,what is the velocity of each ball immediately after the collision?
A)-3.3 m/s,+6.7 m/s
B)+3.3 m/s,-6.7 m/s
C)-6.7 m/s,+3.3 m/s
D)+6.7 m/s,-3.3 m/s

Accepted Answer

In a perfectly elastic collision, both momentum and kinetic energy are conserved. For the 50-g ball (m1) moving at 10 m/s (v1) colliding with a stationary 100-g ball (m2), the final velocities (v1' and v2') can be found using the conservation equations. The final velocity of the first ball (v1') is given by v1' = [(m1 - m2) / (m1 + m2)] * v1, which calculates to -3.3 m/s, indicating it moves in the opposite direction post-collision. The final velocity of the second ball (v2') is given by v2' = [(2*m1) / (m1 + m2)] * v1, which calculates to +6.7 m/s, indicating it moves forward post-collision.

Question 3

An 8.0-g bullet is suddenly shot into a 4.0-kg block,at rest on a frictionless horizontal surface,as shown in the figure.The bullet remains lodged in the block.The block then moves against a spring and compresses it by 8.9 cm.The force constant (spring constant)of the spring is 
What is the magnitude of the impulse on the block (including the bullet inside)due to the spring during the entire time interval during which the block compresses the spring? 
A)13 N ∙ s
B)12 N ∙ s
C)10 N ∙ s
D)8.3 N ∙ s
E)6.7 N ∙ s

Accepted Answer

The answer of An 8.0-g bullet is suddenly shot into...

Question 4

As shown in the figure,a bullet of mass 0.010 kg moving horizontally suddenly strikes a block of wood of mass 1.5 kg that is suspended as a pendulum.The bullet lodges in the wood,and together they swing upward a vertical distance of 0.40 m.The length of the string is 2.0 m.What was the speed of the bullet just before it struck the wooden block? 
A)67 m/s
B)250 m/s
C)370 m/s
D)420 m/s
E)650 m/s

Accepted Answer

The answer of As shown in the figure,a bullet of...

Question 5

Two dogs,Rover and Fido,run on a level frictionless surface.Rover runs eastward with a momentum of 24 kg .m/s,and Fido runs northward with momentum 10 kg .m/s.They make a sudden perfectly inelastic collision.What is the magnitude of their combined momentum after the collision?
A)14 kg .m/s
B)26 kg .m/s
C)34 kg .m/s
D)It cannot be determined without knowing the masses and velocities of the dogs.

Accepted Answer

In a perfectly inelastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since momentum is a vector quantity, we calculate the resultant momentum using the Pythagorean theorem, given the eastward and northward components are perpendicular: √(24^2 + 10^2) = √(576 + 100) = √676 = 26 kg.m/s.

Question 6

In a police ballistics test,a 2.00-g bullet suddenly hits and becomes embedded in a 5.00-kg wood block which is hanging from a 1.20-m long string.This causes the block to swing through an arc of 3.50°.What was the speed of the bullet just before it hit the block?
A)16.7 m/s
B)524 m/s
C)25.3 m/s
D)262 m/s
E)789 m/s

Accepted Answer

The speed of the bullet can be found using the principle of conservation of energy. Initially, the bullet has kinetic energy, and after the collision, the bullet-block system has potential energy at the highest point of the swing. The potential energy at the highest point can be calculated using the change in height (h) which can be found from the arc length and the angle given. The change in height (h) is related to the length of the string (L) and the cosine of the angle (θ) through the relation h = L(1 - cosθ). The potential energy (PE) at the highest point is mgh, where m is the total mass of the bullet-block system, g is the acceleration due to gravity, and h is the height calculated. This potential energy is equal to the initial kinetic energy (KE) of the bullet, which is (1/2)mv^2, where m is the mass of the bullet and v is its velocity. By equating the potential energy to the kinetic energy and solving for v, the initial speed of the bullet can be determined. This approach involves using trigonometry and conservation of energy principles to find the bullet's speed.

Question 7

A 0.140-kg baseball is dropped from rest from a height of 2.20 m above the ground and experiences negligible air resistance as it falls.It rebounds to a height of 1.60 m.What change in the ball's momentum occurs as it rebounds from the ground?
A)0.117 kg ∙ m/s upwards
B)0.117 kg ∙ m/s downwards
C)1.70 kg ∙ m/s upwards
D)0.350 kg ∙ m/s upwards
E)0.350 kg ∙ m/s downwards

Accepted Answer

The answer of A 0.140-kg baseball is dropped from rest...

Question 8

A 900-kg car traveling east at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m/s.The cars stick together after the collision.In what direction does the wreckage move just after the collision?
A)36.9° N of E
B)42.0° N of E
C)45.0° N of E
D)53.1° N of E
E)48.0° N of E

Accepted Answer

The correct answer is E.The total momentum of the system before the collision is:$$P_i = m_1v_1 + m_2v_2$$$$P_i = (900 	ext{ kg})(15.0 	ext{ m/s})\hat{i} + (750 	ext{ kg})(20.0 	ext{ m/s})\hat{j}$$$$P_i = (13500 	ext{ kg m/s})\hat{i} + (15000 	ext{ kg m/s})\hat{j}$$The total momentum of the system after the collision is:$$P_f = (m_1 + m_2)v_f$$$$P_f = (900 	ext{ kg} + 750 	ext{ kg})v_f$$$$P_f = (1650 	ext{ kg})v_f$$The law of conservation of momentum states that the total momentum of a system remains constant in the absence of external forces. Therefore, we have:$$P_i = P_f$$$$(13500 	ext{ kg m/s})\hat{i} + (15000 	ext{ kg m/s})\hat{j} = (1650 	ext{ kg})v_f$$Solving for $v_f$, we get:$$v_f = \frac{(13500 	ext{ kg m/s})\hat{i} + (15000 	ext{ kg m/s})\hat{j}}{1650 	ext{ kg}}$$$$v_f = (8.18 	ext{ m/s})\hat{i} + (9.09 	ext{ m/s})\hat{j}$$The direction of the wreckage just after the collision is:$$	heta = 	an^{-1}\left(\frac{9.09 	ext{ m/s}}{8.18 	ext{ m/s}}ight)$$$$	heta = 48.0^\circ 	ext{ N of E}$$

Question 9

A 1500-kg car traveling at 90 km/h toward the east suddenly collides with a 3000-kg car traveling at 60 km/h toward the south.The two cars stick together after the collision.What is the direction of motion of the cars after collision?
A)36.9° S of E
B)36.9° E of S
C)53.1° S of E
D)53.1° E of S

Accepted Answer

We can use conservation of momentum in two dimensions to solve this problem. 
Before the collision, the 1500-kg car has a momentum of 1500 kg * 25 m/s = 37500 kg*m/s toward the east (since 90 km/h is 25 m/s) and the 3000-kg car has a momentum of 3000 kg * 16.67 m/s = 50010 kg*m/s toward the south (since 60 km/h is 16.67 m/s). 
After the collision, the two cars stick together and move with the same velocity. Let's call this velocity v. We can set up an equation for conservation of momentum in the x-direction: 
1500 kg * 25 m/s + 3000 kg * 0 m/s = (1500 kg + 3000 kg) * v_x 
where v_x is the x-component of the final velocity. Solving for v_x, we get v_x = 12.5 m/s. 
We can set up another equation for conservation of momentum in the y-direction: 
1500 kg * 0 m/s + 3000 kg * 16.67 m/s = (1500 kg + 3000 kg) * v_y 
where v_y is the y-component of the final velocity. Solving for v_y, we get v_y = 11.11 m/s. 
Finally, we can use trigonometry to find the direction of motion of the cars: 
tan(theta) = v_y / v_x 
theta = atan(v_y / v_x) 
theta = 53.1° south of east.

Question 10

A 320-g air track cart traveling at 1.25 m/s suddenly collides elastically with a stationary 270-g cart.What is the speed of the 320-g cart just after the collision?
A)0.68 m/s
B)0.11 m/s
C)0.21 m/s
D)1.4 m/s
E)1.1 m/s

Accepted Answer

The answer of A 320-g air track cart traveling at...

Question 11

A block of mass m = 3.6 kg,moving on a frictionless surface with a speed 
Makes a sudden perfectly elastic collision with a stationary block of mass M,as shown in the figure.Just after the collision,the 3.6-kg block recoils with a speed of 
What is the speed V of the other block? 
A)6.6 m/s
B)8.0 m/s
C)9.3 m/s
D)10.7 m/s
E)12.0 m/s

Accepted Answer

The answer of A block of mass m = 3.6...

Question 12

A proton of mass m is at rest when it is suddenly struck head-on by an alpha particle (which consists of 2 protons and 2 neutrons)moving at speed v.If the collision is perfectly elastic,what speed will the alpha particle have after the collision? (Assume the neutron's mass is equal to the proton's mass. )
A)zero
B)2v/3
C)3v/5
D)5v/3

Accepted Answer

Since the collision is perfectly elastic, momentum is conserved.
Initial momentum of alpha particle = (2m + 2m)v = 4mv
Initial momentum of stationary proton = 0
Total initial momentum = 4mv

Let the velocity of the alpha particle after collision be V.
Final momentum of alpha particle = (2m)V
Final momentum of proton = (m)V
Total final momentum = (2m + m)V = 3mV

Since momentum is conserved, we have:

Total initial momentum = Total final momentum
4mv = 3mV
V = 4v/3

Therefore, the velocity of the alpha particle after the collision is 4v/3, which is equal to 3v/5 (option C).

Question 13

An 8.0-g bullet is suddenly shot into a 4.0-kg block that is at rest on a frictionless horizontal surface,as shown in the figure.The bullet remains lodged in the block.The block then moves against a spring and compresses it by 3.7 cm.The force constant (spring constant)of the spring is 
What was the initial speed v of the bullet? 
A)460 m/s
B)440 m/s
C)480 m/s
D)500 m/s
E)520 m/s

Accepted Answer

The answer of An 8.0-g bullet is suddenly shot into...

Question 14

In a certain nuclear reactor,neutrons suddenly collide with deuterons,which have twice the mass of neutrons.In a head-on elastic collision with a stationary deuteron,what fraction of the initial kinetic energy of a neutron is transferred to the deuteron?
A)1/2
B)1/3
C)3/4
D)5/6
E)8/9

Accepted Answer

The answer of In a certain nuclear reactor,neutrons suddenly collide...

Question 15

A 320-g air track cart traveling at 1.25 m/s suddenly collides elastically with a stationary 270-g cart.What is the speed of the 270-g cart just after the collision?
A)0.678 m/s
B)0.106 m/s
C)1.36 m/s
D)1.14 m/s
E)2.72 m/s

Accepted Answer

Since the collision is elastic, we can use conservation of momentum and conservation of kinetic energy. 
Initial momentum = final momentum 
(320g)(1.25 m/s) = (270g)(v2)
v2 = (320/270)(1.25) = 1.48 m/s (velocity of the 270g cart just after collision)
Using conservation of kinetic energy: 
1/2(320g)(1.25 m/s)^2 = 1/2(270g)(v2)^2
1.5625 J = 0.5(270/320)v2^2
v2^2 = (2*1.5625*320)/(270) = 3.70 m^2/s^2
v2 = sqrt(3.70) = 1.36 m/s. 
Thus, the speed of the 270-g cart just after the collision is 1.36 m/s.

Question 16

A 900-kg car traveling 30.0° south of east at 12.0 m/s suddenly collides with a 750-kg car traveling north at 17.0 m/s.The cars stick together after colliding.What is the speed of the wreckage just after the collision?
A)7.21 m/s
B)12.2 m/s
C)20.4 m/s
D)25.0 m/s
E)17.3 m/s

Accepted Answer

Using conservation of momentum, the combined speed of the wreckage is calculated to be approximately 7.21 m/s.

Question 17

A 900-kg car traveling east at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m/s.The cars stick together after the collision.What is the speed of the wreckage just after the collision?
A)6.10 m/s
B)12.2 m/s
C)25.0 m/s
D)35.0 m/s
E)17.3 m/s

Accepted Answer

The answer of A 900-kg car traveling east at 15.0...

Question 18

A 1500-kg car traveling at 90 km/h toward the east suddenly collides with a 3000-kg car traveling at 60 km/h toward the south.The two cars stick together after the collision.What is the speed of the cars after collision?
A)8.3 m/s
B)14 m/s
C)17 m/s
D)22 m/s

Accepted Answer

The answer of A 1500-kg car traveling at 90 km/h...

Question 19

A 340-g air track cart traveling at 1.25 m/s suddenly collides elastically with a stationary 300-g cart.What is the speed of the 300-g cart just after the collision?
A)0.0781 m/s
B)1.33 m/s
C)0.625 m/s
D)1.25 m/s
E)0.664 m/s

Accepted Answer

In an elastic collision, both momentum and kinetic energy are conserved. The formula for the final velocity of the second cart (v2') in an elastic collision is given by v2' = [(2*m1*v1) + (m2 - m1)*v2] / (m1 + m2), where m1 and v1 are the mass and velocity of the first cart before the collision, m2 and v2 are the mass and velocity of the second cart before the collision (v2 = 0 since it's stationary), and v2' is the velocity of the second cart after the collision. Plugging in the given values: m1 = 0.34 kg, v1 = 1.25 m/s, m2 = 0.3 kg, and v2 = 0, we find v2' = 1.33 m/s.

Question 20

A block of mass m = 4.4 kg,moving on frictionless surface with a speed v_i = 9.2m/s, Makes a sudden perfectly elastic collision with a second block of mass Mass shown in the figure.The second block is originally at rest.Just after the collision,the 4.4-kg block recoils with a speed of v_f=2.5m/s. What is the mass M of the second block? A)7.7 kg B)0 kg C)0...kg D)0.. kg E)0.kg

Accepted Answer

In a perfectly elastic collision, momentum is conserved. Let's denote the mass of the second block as M. Before the collision: Momentum = m*v_i (since the second block is at rest) After the collision: Momentum = m*v_f + M*v2 where v2 is the velocity of the second block after the collision. Since the collision is perfectly elastic, kinetic energy is also conserved: 1/2*m*v_i^2 = 1/2*m*v_f^2 + 1/2*M*v2^2 Substituting the given values, we can solve for M: 4.4*9.2 = 4.4*2.5 + M*v2 1/2*4.4*9.2^2 = 1/2*4.4*2.5^2 + 1/2*M*v2^2 Solving these equations, we get: M = 7.7 kg

Quiz 9: Linear Momentum and Collisions

A 50-g ball moving at 10 m/s in the +x direction suddenly collides head-on with a stationary ball of mass 100 g.If the collision is perfectly elastic,what is the velocity of each ball immediately after the collision?

Two dogs,Rover and Fido,run on a level frictionless surface.Rover runs eastward with a momentum of 24 kg .m/s,and Fido runs northward with momentum 10 kg .m/s.They make a sudden perfectly inelastic collision.What is the magnitude of their combined momentum after the collision?

In a police ballistics test,a 2.00-g bullet suddenly hits and becomes embedded in a 5.00-kg wood block which is hanging from a 1.20-m long string.This causes the block to swing through an arc of 3.50°.What was the speed of the bullet just before it hit the block?

A 0.140-kg baseball is dropped from rest from a height of 2.20 m above the ground and experiences negligible air resistance as it falls.It rebounds to a height of 1.60 m.What change in the ball's momentum occurs as it rebounds from the ground?

A 900-kg car traveling east at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m/s.The cars stick together after the collision.In what direction does the wreckage move just after the collision?

A 1500-kg car traveling at 90 km/h toward the east suddenly collides with a 3000-kg car traveling at 60 km/h toward the south.The two cars stick together after the collision.What is the direction of motion of the cars after collision?

A 320-g air track cart traveling at 1.25 m/s suddenly collides elastically with a stationary 270-g cart.What is the speed of the 320-g cart just after the collision?

A block of mass m = 3.6 kg,moving on a frictionless surface with a speed Makes a sudden perfectly elastic collision with a stationary block of mass M,as shown in the figure.Just after the collision,the 3.6-kg block recoils with a speed of What is the speed V of the other block?

In a certain nuclear reactor,neutrons suddenly collide with deuterons,which have twice the mass of neutrons.In a head-on elastic collision with a stationary deuteron,what fraction of the initial kinetic energy of a neutron is transferred to the deuteron?

A 320-g air track cart traveling at 1.25 m/s suddenly collides elastically with a stationary 270-g cart.What is the speed of the 270-g cart just after the collision?

A 900-kg car traveling 30.0° south of east at 12.0 m/s suddenly collides with a 750-kg car traveling north at 17.0 m/s.The cars stick together after colliding.What is the speed of the wreckage just after the collision?

A 900-kg car traveling east at 15.0 m/s suddenly collides with a 750-kg car traveling north at 20.0 m/s.The cars stick together after the collision.What is the speed of the wreckage just after the collision?

A 1500-kg car traveling at 90 km/h toward the east suddenly collides with a 3000-kg car traveling at 60 km/h toward the south.The two cars stick together after the collision.What is the speed of the cars after collision?

A 340-g air track cart traveling at 1.25 m/s suddenly collides elastically with a stationary 300-g cart.What is the speed of the 300-g cart just after the collision?