A block with mass m1= .500kg is released from rest on a frictionless track at a

Question

A block with mass m1= .500kg is released from rest on a frictionless track at a distance h1 = 2.5m above the top of the table. It then collides elastically with a object having a mass m2= 1.00 kg that is initially at rest on the table. A) determine the velocities of the 2 objects just after collision. B) How high up thetrack does the .500 kg object travel after the collision? C) How far from the bottom of the table does the 1.00 kg object land, given the height of the table is h2 = 2.00m? D) How far away from the bottom of the table does the .500kg object eventually land?

Explanation / Answer

A m1 = 0.550 kg block is released from rest at the top of a frictionless track h1 = 2.65 m above the top of a table. It then collides elastically with a 1.00 kg block that is initially at rest on the table, as shown in Figure P6.57.

http://www.webassign.net/sf/p6_57alt.gif

(a) Determine the velocities of the two objects just after the collision.
velocity of m1:
velocity of m2:

(b) How high up the track does the 0.550 kg object travel back after the collision?

(c) How far away from the bottom of the table does the 1.00 kg object land, given that the table is 2.20 m high?

(d) How far away from the bottom of the table does the 0.550 kg object eventually land?

• ANS

A m1 = 0.550 kg block is released from rest at the top of a frictionless track h1 = 2.65 m above the top of a table. It then collides elastically with a 1.00 kg block that is initially at rest on the table, as shown in Figure P6.57.

http://www.webassign.net/sf/p6_57alt.gif

(a) Determine the velocities of the two objects just after the collision.
velocity of m1:
velocity of m2:

(b) How high up the track does the 0.550 kg object travel back after the collision?

(c) How far away from the bottom of the table does the 1.00 kg object land, given that the table is 2.20 m high?

(d) How far away from the bottom of the table does the 0.550 kg object eventually land?

• ANS

A m1 = 0.550 kg block is released from rest at the top of a frictionless track h1 = 2.65 m above the top of a table. It then collides elastically with a 1.00 kg block that is initially at rest on the table, as shown in Figure P6.57.

http://www.webassign.net/sf/p6_57alt.gif

(a) Determine the velocities of the two objects just after the collision.
velocity of m1:
velocity of m2:

(b) How high up the track does the 0.550 kg object travel back after the collision?

(c) How far away from the bottom of the table does the 1.00 kg object land, given that the table is 2.20 m high?

(d) How far away from the bottom of the table does the 0.550 kg object eventually land?

a) The velocity of m1 just before impact is found from
Pe=Ke
m1gh=(1/2) m1 U1^2
U1=sqrt(2 gh)=7.2m/s

Conservation of momentum
m1U1 +m2U2=m1V1 + m2V2
It helps to know that U2= 0

m1U1 = m1V1 + m2V2
also
What about conservation of kinetic energy? Ke=(1/2) mV^2
Ke1+Ke2=Ke1' + Ke2' again Ke2=0 and 1/2 term can be dropped

m1U1^2= m1V1^2 + m2V2^2 (See ref 1)
solving these equations we get

V1=(m1-m2)U1/(m1+m2) and since U1=sqrt(2 gh)= 7.2 m/s
V1=(0.550 - 1.00) (7.2)/(550 +1.00)
V1=-2.1m/s

V2= 2m1U1/(m1+m2)

b) Again Ke(1/2)mV^2=mgh
h=(1/2)V^2 /g= 0.5 x (2.1)^2 /9.81=.222 m

c) Time of fall =time of horizontal travel. Horizontal velocity is constant
t=sqrt(2 h2 / g)
t= sqrt( 2 x 2.2 / 9.81)=0.67 s
S= V2 t= 2m1U1/(m1+m2) t=
S=[2 x 0.550 x 7.2/(0.55 +1.00) ] 0.67 =
S=3.42m

d) The horizontal velocity will be the same in magnitude as it was after m1 colided with m2. The time of flight is the sam efor both of them so

S= V1t= 2.1 x 0.67= 1.41 m

Have fun!

a) The velocity of m1 just before impact is found from
Pe=Ke
m1gh=(1/2) m1 U1^2
U1=sqrt(2 gh)=7.2m/s

Conservation of momentum
m1U1 +m2U2=m1V1 + m2V2
It helps to know that U2= 0

m1U1 = m1V1 + m2V2
also
What about conservation of kinetic energy? Ke=(1/2) mV^2
Ke1+Ke2=Ke1' + Ke2' again Ke2=0 and 1/2 term can be dropped

m1U1^2= m1V1^2 + m2V2^2 (See ref 1)
solving these equations we get

V1=(m1-m2)U1/(m1+m2) and since U1=sqrt(2 gh)= 7.2 m/s
V1=(0.550 - 1.00) (7.2)/(550 +1.00)
V1=-2.1m/s

V2= 2m1U1/(m1+m2)

b) Again Ke(1/2)mV^2=mgh
h=(1/2)V^2 /g= 0.5 x (2.1)^2 /9.81=.222 m

c) Time of fall =time of horizontal travel. Horizontal velocity is constant
t=sqrt(2 h2 / g)
t= sqrt( 2 x 2.2 / 9.81)=0.67 s
S= V2 t= 2m1U1/(m1+m2) t=
S=[2 x 0.550 x 7.2/(0.55 +1.00) ] 0.67 =
S=3.42m

d) The horizontal velocity will be the same in magnitude as it was after m1 colided with m2. The time of flight is the sam efor both of them so

S= V1t= 2.1 x 0.67= 1.41 m

Have fun!

a) The velocity of m1 just before impact is found from
Pe=Ke
m1gh=(1/2) m1 U1^2
U1=sqrt(2 gh)=7.2m/s

Conservation of momentum
m1U1 +m2U2=m1V1 + m2V2
It helps to know that U2= 0

m1U1 = m1V1 + m2V2
also
What about conservation of kinetic energy? Ke=(1/2) mV^2
Ke1+Ke2=Ke1' + Ke2' again Ke2=0 and 1/2 term can be dropped

m1U1^2= m1V1^2 + m2V2^2 (See ref 1)
solving these equations we get

V1=(m1-m2)U1/(m1+m2) and since U1=sqrt(2 gh)= 7.2 m/s
V1=(0.550 - 1.00) (7.2)/(550 +1.00)
V1=-2.1m/s

V2= 2m1U1/(m1+m2)

b) Again Ke(1/2)mV^2=mgh
h=(1/2)V^2 /g= 0.5 x (2.1)^2 /9.81=.222 m

c) Time of fall =time of horizontal travel. Horizontal velocity is constant
t=sqrt(2 h2 / g)
t= sqrt( 2 x 2.2 / 9.81)=0.67 s
S= V2 t= 2m1U1/(m1+m2) t=
S=[2 x 0.550 x 7.2/(0.55 +1.00) ] 0.67 =
S=3.42m

d) The horizontal velocity will be the same in magnitude as it was after m1 colided with m2. The time of flight is the sam efor both of them so

S= V1t= 2.1 x 0.67= 1.41 m

Have fun!

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.